Shared Qs (027)

1. 
   

   Question

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   In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. The simplest version of partial-fraction decomposition occurs when there are two distinct linear factors.

   The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restrictions \(x\ne -7\) and \(x\ne 2\)).

   \[\frac{3x-33}{(x+7)(x-2)} = \frac{A}{x+7}+\frac{B}{x-2}\]

   • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   
   
   
   

   Solution

   Show solution

   We are trying to find \(A\) and \(B\), to make both sides of the equation equivalent expressions.

   \[\frac{3x-33}{(x+7)(x-2)} ~=~ \frac{A}{x+7}+\frac{B}{x-2}\]

   Multiply both sides by \((x+7)(x-2)\).

   \[\frac{3x-33}{(x+7)(x-2)} \cdot (x+7)(x-2) ~=~ \left(\frac{A}{x+7}+\frac{B}{x-2}\right) \cdot (x+7)(x-2)\]

   Simplify.

   \[3x-33 = A(x-2)+B(x+7)\]

   Expand the right side of the equation. Gather the linear coefficient and the constant.

   \[3x-33 = (A+B)x+(-2A+7B)\] The equation must be true FOR ALL values of \(x\). This means the linear coefficients must be equal, and the constants must be equal.

   \[\begin{align} 3 &= A + B \\ -33 &= -2A+7B\\ \end{align}\]

   Use elimination. Multiply the first equation by 2.

   \[\begin{align} 6 &= 2A+2B \\ \oplus~~~~ -33 &= -2A+7B\\ \hline -27 &= 9B \end{align}\]

   Divide both sides by 9.

   \[-3 = B\]

   Plug that value into \(3 = A + B\).

   \[3 = A-3\]

   Add 3 to both sides.

   \[6 = A\]

   
   
2. 
   

   Question

   prevnext

   In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. The simplest version of partial-fraction decomposition occurs when there are two distinct linear factors.

   The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restrictions \(x\ne -5\) and \(x\ne -8\)).

   \[\frac{5x+46}{(x+5)(x+8)} = \frac{A}{x+5}+\frac{B}{x+8}\]

   • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   
   
   
   

   Solution

   Show solution

   We are trying to find \(A\) and \(B\), to make both sides of the equation equivalent expressions.

   \[\frac{5x+46}{(x+5)(x+8)} ~=~ \frac{A}{x+5}+\frac{B}{x+8}\]

   Multiply both sides by \((x+5)(x+8)\).

   \[\frac{5x+46}{(x+5)(x+8)} \cdot (x+5)(x+8) ~=~ \left(\frac{A}{x+5}+\frac{B}{x+8}\right) \cdot (x+5)(x+8)\]

   Simplify.

   \[5x+46 = A(x+8)+B(x+5)\]

   Expand the right side of the equation. Gather the linear coefficient and the constant.

   \[5x+46 = (A+B)x+(8A+5B)\] The equation must be true FOR ALL values of \(x\). This means the linear coefficients must be equal, and the constants must be equal.

   \[\begin{align} 5 &= A + B \\ 46 &= 8A+5B\\ \end{align}\]

   Use elimination. Multiply the first equation by -8.

   \[\begin{align} -40 &= -8A-8B \\ \oplus~~~~ 46 &= 8A+5B\\ \hline 6 &= -3B \end{align}\]

   Divide both sides by -3.

   \[-2 = B\]

   Plug that value into \(5 = A + B\).

   \[5 = A-2\]

   Add 2 to both sides.

   \[7 = A\]

   
   
3. 
   

   Question

   prevnext

   In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. The simplest version of partial-fraction decomposition occurs when there are two distinct linear factors.

   The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restrictions \(x\ne 2\) and \(x\ne -9\)).

   \[\frac{-12x-64}{(x-2)(x+9)} = \frac{A}{x-2}+\frac{B}{x+9}\]

   • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   
   
   
   

   Solution

   Show solution

   We are trying to find \(A\) and \(B\), to make both sides of the equation equivalent expressions.

   \[\frac{-12x-64}{(x-2)(x+9)} ~=~ \frac{A}{x-2}+\frac{B}{x+9}\]

   Multiply both sides by \((x-2)(x+9)\).

   \[\frac{-12x-64}{(x-2)(x+9)} \cdot (x-2)(x+9) ~=~ \left(\frac{A}{x-2}+\frac{B}{x+9}\right) \cdot (x-2)(x+9)\]

   Simplify.

   \[-12x-64 = A(x+9)+B(x-2)\]

   Expand the right side of the equation. Gather the linear coefficient and the constant.

   \[-12x-64 = (A+B)x+(9A-2B)\] The equation must be true FOR ALL values of \(x\). This means the linear coefficients must be equal, and the constants must be equal.

   \[\begin{align} -12 &= A + B \\ -64 &= 9A-2B\\ \end{align}\]

   Use elimination. Multiply the first equation by -9.

   \[\begin{align} 108 &= -9A-9B \\ \oplus~~~~ -64 &= 9A-2B\\ \hline 44 &= -11B \end{align}\]

   Divide both sides by -11.

   \[-4 = B\]

   Plug that value into \(-12 = A + B\).

   \[-12 = A-4\]

   Add 4 to both sides.

   \[-8 = A\]

   
   
4. 
   

   Question

   prevnext

   In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. The simplest version of partial-fraction decomposition occurs when there are two distinct linear factors.

   The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restrictions \(x\ne 6\) and \(x\ne -2\)).

   \[\frac{13x-38}{(x-6)(x+2)} = \frac{A}{x-6}+\frac{B}{x+2}\]

   • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   
   
   
   

   Solution

   Show solution

   We are trying to find \(A\) and \(B\), to make both sides of the equation equivalent expressions.

   \[\frac{13x-38}{(x-6)(x+2)} ~=~ \frac{A}{x-6}+\frac{B}{x+2}\]

   Multiply both sides by \((x-6)(x+2)\).

   \[\frac{13x-38}{(x-6)(x+2)} \cdot (x-6)(x+2) ~=~ \left(\frac{A}{x-6}+\frac{B}{x+2}\right) \cdot (x-6)(x+2)\]

   Simplify.

   \[13x-38 = A(x+2)+B(x-6)\]

   Expand the right side of the equation. Gather the linear coefficient and the constant.

   \[13x-38 = (A+B)x+(2A-6B)\] The equation must be true FOR ALL values of \(x\). This means the linear coefficients must be equal, and the constants must be equal.

   \[\begin{align} 13 &= A + B \\ -38 &= 2A-6B\\ \end{align}\]

   Use elimination. Multiply the first equation by -2.

   \[\begin{align} -26 &= -2A-2B \\ \oplus~~~~ -38 &= 2A-6B\\ \hline -64 &= -8B \end{align}\]

   Divide both sides by -8.

   \[8 = B\]

   Plug that value into \(13 = A + B\).

   \[13 = A+8\]

   Add -8 to both sides.

   \[5 = A\]

   
   
5. 
   

   Question

   prevnext

   In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. The simplest version of partial-fraction decomposition occurs when there are two distinct linear factors.

   The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restrictions \(x\ne 3\) and \(x\ne -7\)).

   \[\frac{-7x-29}{(x-3)(x+7)} = \frac{A}{x-3}+\frac{B}{x+7}\]

   • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   
   
   
   

   Solution

   Show solution

   We are trying to find \(A\) and \(B\), to make both sides of the equation equivalent expressions.

   \[\frac{-7x-29}{(x-3)(x+7)} ~=~ \frac{A}{x-3}+\frac{B}{x+7}\]

   Multiply both sides by \((x-3)(x+7)\).

   \[\frac{-7x-29}{(x-3)(x+7)} \cdot (x-3)(x+7) ~=~ \left(\frac{A}{x-3}+\frac{B}{x+7}\right) \cdot (x-3)(x+7)\]

   Simplify.

   \[-7x-29 = A(x+7)+B(x-3)\]

   Expand the right side of the equation. Gather the linear coefficient and the constant.

   \[-7x-29 = (A+B)x+(7A-3B)\] The equation must be true FOR ALL values of \(x\). This means the linear coefficients must be equal, and the constants must be equal.

   \[\begin{align} -7 &= A + B \\ -29 &= 7A-3B\\ \end{align}\]

   Use elimination. Multiply the first equation by -7.

   \[\begin{align} 49 &= -7A-7B \\ \oplus~~~~ -29 &= 7A-3B\\ \hline 20 &= -10B \end{align}\]

   Divide both sides by -10.

   \[-2 = B\]

   Plug that value into \(-7 = A + B\).

   \[-7 = A-2\]

   Add 2 to both sides.

   \[-5 = A\]

   
   
6. 
   

   Question

   prevnext

   In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. The simplest version of partial-fraction decomposition occurs when there are two distinct linear factors.

   The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restrictions \(x\ne -6\) and \(x\ne -9\)).

   \[\frac{9x+60}{(x+6)(x+9)} = \frac{A}{x+6}+\frac{B}{x+9}\]

   • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   
   
   
   

   Solution

   Show solution

   We are trying to find \(A\) and \(B\), to make both sides of the equation equivalent expressions.

   \[\frac{9x+60}{(x+6)(x+9)} ~=~ \frac{A}{x+6}+\frac{B}{x+9}\]

   Multiply both sides by \((x+6)(x+9)\).

   \[\frac{9x+60}{(x+6)(x+9)} \cdot (x+6)(x+9) ~=~ \left(\frac{A}{x+6}+\frac{B}{x+9}\right) \cdot (x+6)(x+9)\]

   Simplify.

   \[9x+60 = A(x+9)+B(x+6)\]

   Expand the right side of the equation. Gather the linear coefficient and the constant.

   \[9x+60 = (A+B)x+(9A+6B)\] The equation must be true FOR ALL values of \(x\). This means the linear coefficients must be equal, and the constants must be equal.

   \[\begin{align} 9 &= A + B \\ 60 &= 9A+6B\\ \end{align}\]

   Use elimination. Multiply the first equation by -9.

   \[\begin{align} -81 &= -9A-9B \\ \oplus~~~~ 60 &= 9A+6B\\ \hline -21 &= -3B \end{align}\]

   Divide both sides by -3.

   \[7 = B\]

   Plug that value into \(9 = A + B\).

   \[9 = A+7\]

   Add -7 to both sides.

   \[2 = A\]

   
   
7. 
   

   Question

   prevnext

   In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. The simplest version of partial-fraction decomposition occurs when there are two distinct linear factors.

   The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restrictions \(x\ne 2\) and \(x\ne 6\)).

   \[\frac{13x-42}{(x-2)(x-6)} = \frac{A}{x-2}+\frac{B}{x-6}\]

   • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   
   
   
   

   Solution

   Show solution

   We are trying to find \(A\) and \(B\), to make both sides of the equation equivalent expressions.

   \[\frac{13x-42}{(x-2)(x-6)} ~=~ \frac{A}{x-2}+\frac{B}{x-6}\]

   Multiply both sides by \((x-2)(x-6)\).

   \[\frac{13x-42}{(x-2)(x-6)} \cdot (x-2)(x-6) ~=~ \left(\frac{A}{x-2}+\frac{B}{x-6}\right) \cdot (x-2)(x-6)\]

   Simplify.

   \[13x-42 = A(x-6)+B(x-2)\]

   Expand the right side of the equation. Gather the linear coefficient and the constant.

   \[13x-42 = (A+B)x+(-6A-2B)\] The equation must be true FOR ALL values of \(x\). This means the linear coefficients must be equal, and the constants must be equal.

   \[\begin{align} 13 &= A + B \\ -42 &= -6A-2B\\ \end{align}\]

   Use elimination. Multiply the first equation by 6.

   \[\begin{align} 78 &= 6A+6B \\ \oplus~~~~ -42 &= -6A-2B\\ \hline 36 &= 4B \end{align}\]

   Divide both sides by 4.

   \[9 = B\]

   Plug that value into \(13 = A + B\).

   \[13 = A+9\]

   Add -9 to both sides.

   \[4 = A\]

   
   
8. 
   

   Question

   prevnext

   In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. The simplest version of partial-fraction decomposition occurs when there are two distinct linear factors.

   The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restrictions \(x\ne -8\) and \(x\ne 3\)).

   \[\frac{13x+60}{(x+8)(x-3)} = \frac{A}{x+8}+\frac{B}{x-3}\]

   • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   
   
   
   

   Solution

   Show solution

   We are trying to find \(A\) and \(B\), to make both sides of the equation equivalent expressions.

   \[\frac{13x+60}{(x+8)(x-3)} ~=~ \frac{A}{x+8}+\frac{B}{x-3}\]

   Multiply both sides by \((x+8)(x-3)\).

   \[\frac{13x+60}{(x+8)(x-3)} \cdot (x+8)(x-3) ~=~ \left(\frac{A}{x+8}+\frac{B}{x-3}\right) \cdot (x+8)(x-3)\]

   Simplify.

   \[13x+60 = A(x-3)+B(x+8)\]

   Expand the right side of the equation. Gather the linear coefficient and the constant.

   \[13x+60 = (A+B)x+(-3A+8B)\] The equation must be true FOR ALL values of \(x\). This means the linear coefficients must be equal, and the constants must be equal.

   \[\begin{align} 13 &= A + B \\ 60 &= -3A+8B\\ \end{align}\]

   Use elimination. Multiply the first equation by 3.

   \[\begin{align} 39 &= 3A+3B \\ \oplus~~~~ 60 &= -3A+8B\\ \hline 99 &= 11B \end{align}\]

   Divide both sides by 11.

   \[9 = B\]

   Plug that value into \(13 = A + B\).

   \[13 = A+9\]

   Add -9 to both sides.

   \[4 = A\]

   
   
9. 
   

   Question

   prevnext

   In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. The simplest version of partial-fraction decomposition occurs when there are two distinct linear factors.

   The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restrictions \(x\ne 9\) and \(x\ne 5\)).

   \[\frac{-4x+52}{(x-9)(x-5)} = \frac{A}{x-9}+\frac{B}{x-5}\]

   • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
   
   
   
   

   Solution

   Show solution

   We are trying to find \(A\) and \(B\), to make both sides of the equation equivalent expressions.

   \[\frac{-4x+52}{(x-9)(x-5)} ~=~ \frac{A}{x-9}+\frac{B}{x-5}\]

   Multiply both sides by \((x-9)(x-5)\).

   \[\frac{-4x+52}{(x-9)(x-5)} \cdot (x-9)(x-5) ~=~ \left(\frac{A}{x-9}+\frac{B}{x-5}\right) \cdot (x-9)(x-5)\]

   Simplify.

   \[-4x+52 = A(x-5)+B(x-9)\]

   Expand the right side of the equation. Gather the linear coefficient and the constant.

   \[-4x+52 = (A+B)x+(-5A-9B)\] The equation must be true FOR ALL values of \(x\). This means the linear coefficients must be equal, and the constants must be equal.

   \[\begin{align} -4 &= A + B \\ 52 &= -5A-9B\\ \end{align}\]

   Use elimination. Multiply the first equation by 5.

   \[\begin{align} -20 &= 5A+5B \\ \oplus~~~~ 52 &= -5A-9B\\ \hline 32 &= -4B \end{align}\]

   Divide both sides by -4.

   \[-8 = B\]

   Plug that value into \(-4 = A + B\).

   \[-4 = A-8\]

   Add 8 to both sides.

   \[4 = A\]

   
   
10. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. The simplest version of partial-fraction decomposition occurs when there are two distinct linear factors.

    The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restrictions \(x\ne -5\) and \(x\ne -2\)).

    \[\frac{-15x-51}{(x+5)(x+2)} = \frac{A}{x+5}+\frac{B}{x+2}\]

    • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    We are trying to find \(A\) and \(B\), to make both sides of the equation equivalent expressions.

    \[\frac{-15x-51}{(x+5)(x+2)} ~=~ \frac{A}{x+5}+\frac{B}{x+2}\]

    Multiply both sides by \((x+5)(x+2)\).

    \[\frac{-15x-51}{(x+5)(x+2)} \cdot (x+5)(x+2) ~=~ \left(\frac{A}{x+5}+\frac{B}{x+2}\right) \cdot (x+5)(x+2)\]

    Simplify.

    \[-15x-51 = A(x+2)+B(x+5)\]

    Expand the right side of the equation. Gather the linear coefficient and the constant.

    \[-15x-51 = (A+B)x+(2A+5B)\] The equation must be true FOR ALL values of \(x\). This means the linear coefficients must be equal, and the constants must be equal.

    \[\begin{align} -15 &= A + B \\ -51 &= 2A+5B\\ \end{align}\]

    Use elimination. Multiply the first equation by -2.

    \[\begin{align} 30 &= -2A-2B \\ \oplus~~~~ -51 &= 2A+5B\\ \hline -21 &= 3B \end{align}\]

    Divide both sides by 3.

    \[-7 = B\]

    Plug that value into \(-15 = A + B\).

    \[-15 = A-7\]

    Add 7 to both sides.

    \[-8 = A\]

    
    
11. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. One version of partial-fraction decomposition occurs when there is a squared linear factor.

    The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restriction \(x\ne 3\)). It should be said that knowing/remembering how to set up the equation below is the hardest part of partial-fraction decomposition.

    \[\frac{-4x+18}{(x-3)^2} ~=~ \frac{A}{(x-3)}+\frac{B}{(x-3)^2}\]

    • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    \[\frac{-4x+18}{(x-3)^2} ~=~ \frac{A}{(x-3)}+\frac{B}{(x-3)^2}\]

    Multiply both sides by \((x-3)^2\).

    \[-4x+18 ~=~ A(x-3)+B\]

    Expand. Recognize the linear coefficient and the constant.

    \[-4x+18 ~=~ Ax+(-3A+B)\]

    The corresponding coefficients must be equal for the equation to be true for all \(x\).

    \[\begin{align} -4 ~&=~ A \\ 18 ~&=~ -3A+B \end{align}\]

    We already have \(A\). Substitute to get \(B\). \[18 ~=~ -3(-4)+B\]

    Multiply -3 and -4.

    \[18 ~=~ 12+B\]

    Add \(-12\) to both sides.

    \[6 ~=~ B\]

    
    
12. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. One version of partial-fraction decomposition occurs when there is a squared linear factor.

    The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restriction \(x\ne 9\)). It should be said that knowing/remembering how to set up the equation below is the hardest part of partial-fraction decomposition.

    \[\frac{3x-31}{(x-9)^2} ~=~ \frac{A}{(x-9)}+\frac{B}{(x-9)^2}\]

    • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    \[\frac{3x-31}{(x-9)^2} ~=~ \frac{A}{(x-9)}+\frac{B}{(x-9)^2}\]

    Multiply both sides by \((x-9)^2\).

    \[3x-31 ~=~ A(x-9)+B\]

    Expand. Recognize the linear coefficient and the constant.

    \[3x-31 ~=~ Ax+(-9A+B)\]

    The corresponding coefficients must be equal for the equation to be true for all \(x\).

    \[\begin{align} 3 ~&=~ A \\ -31 ~&=~ -9A+B \end{align}\]

    We already have \(A\). Substitute to get \(B\). \[-31 ~=~ -9(3)+B\]

    Multiply -9 and 3.

    \[-31 ~=~ -27+B\]

    Add \(27\) to both sides.

    \[-4 ~=~ B\]

    
    
13. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. One version of partial-fraction decomposition occurs when there is a squared linear factor.

    The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restriction \(x\ne -7\)). It should be said that knowing/remembering how to set up the equation below is the hardest part of partial-fraction decomposition.

    \[\frac{6x+33}{(x+7)^2} ~=~ \frac{A}{(x+7)}+\frac{B}{(x+7)^2}\]

    • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    \[\frac{6x+33}{(x+7)^2} ~=~ \frac{A}{(x+7)}+\frac{B}{(x+7)^2}\]

    Multiply both sides by \((x+7)^2\).

    \[6x+33 ~=~ A(x+7)+B\]

    Expand. Recognize the linear coefficient and the constant.

    \[6x+33 ~=~ Ax+(7A+B)\]

    The corresponding coefficients must be equal for the equation to be true for all \(x\).

    \[\begin{align} 6 ~&=~ A \\ 33 ~&=~ 7A+B \end{align}\]

    We already have \(A\). Substitute to get \(B\). \[33 ~=~ 7(6)+B\]

    Multiply 7 and 6.

    \[33 ~=~ 42+B\]

    Add \(-42\) to both sides.

    \[-9 ~=~ B\]

    
    
14. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. One version of partial-fraction decomposition occurs when there is a squared linear factor.

    The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restriction \(x\ne 2\)). It should be said that knowing/remembering how to set up the equation below is the hardest part of partial-fraction decomposition.

    \[\frac{3x-2}{(x-2)^2} ~=~ \frac{A}{(x-2)}+\frac{B}{(x-2)^2}\]

    • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    \[\frac{3x-2}{(x-2)^2} ~=~ \frac{A}{(x-2)}+\frac{B}{(x-2)^2}\]

    Multiply both sides by \((x-2)^2\).

    \[3x-2 ~=~ A(x-2)+B\]

    Expand. Recognize the linear coefficient and the constant.

    \[3x-2 ~=~ Ax+(-2A+B)\]

    The corresponding coefficients must be equal for the equation to be true for all \(x\).

    \[\begin{align} 3 ~&=~ A \\ -2 ~&=~ -2A+B \end{align}\]

    We already have \(A\). Substitute to get \(B\). \[-2 ~=~ -2(3)+B\]

    Multiply -2 and 3.

    \[-2 ~=~ -6+B\]

    Add \(6\) to both sides.

    \[4 ~=~ B\]

    
    
15. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. One version of partial-fraction decomposition occurs when there is a squared linear factor.

    The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restriction \(x\ne 7\)). It should be said that knowing/remembering how to set up the equation below is the hardest part of partial-fraction decomposition.

    \[\frac{-2x+10}{(x-7)^2} ~=~ \frac{A}{(x-7)}+\frac{B}{(x-7)^2}\]

    • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    \[\frac{-2x+10}{(x-7)^2} ~=~ \frac{A}{(x-7)}+\frac{B}{(x-7)^2}\]

    Multiply both sides by \((x-7)^2\).

    \[-2x+10 ~=~ A(x-7)+B\]

    Expand. Recognize the linear coefficient and the constant.

    \[-2x+10 ~=~ Ax+(-7A+B)\]

    The corresponding coefficients must be equal for the equation to be true for all \(x\).

    \[\begin{align} -2 ~&=~ A \\ 10 ~&=~ -7A+B \end{align}\]

    We already have \(A\). Substitute to get \(B\). \[10 ~=~ -7(-2)+B\]

    Multiply -7 and -2.

    \[10 ~=~ 14+B\]

    Add \(-14\) to both sides.

    \[-4 ~=~ B\]

    
    
16. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. One version of partial-fraction decomposition occurs when there is a squared linear factor.

    The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restriction \(x\ne 4\)). It should be said that knowing/remembering how to set up the equation below is the hardest part of partial-fraction decomposition.

    \[\frac{-2x+16}{(x-4)^2} ~=~ \frac{A}{(x-4)}+\frac{B}{(x-4)^2}\]

    • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    \[\frac{-2x+16}{(x-4)^2} ~=~ \frac{A}{(x-4)}+\frac{B}{(x-4)^2}\]

    Multiply both sides by \((x-4)^2\).

    \[-2x+16 ~=~ A(x-4)+B\]

    Expand. Recognize the linear coefficient and the constant.

    \[-2x+16 ~=~ Ax+(-4A+B)\]

    The corresponding coefficients must be equal for the equation to be true for all \(x\).

    \[\begin{align} -2 ~&=~ A \\ 16 ~&=~ -4A+B \end{align}\]

    We already have \(A\). Substitute to get \(B\). \[16 ~=~ -4(-2)+B\]

    Multiply -4 and -2.

    \[16 ~=~ 8+B\]

    Add \(-8\) to both sides.

    \[8 ~=~ B\]

    
    
17. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. One version of partial-fraction decomposition occurs when there is a squared linear factor.

    The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restriction \(x\ne -2\)). It should be said that knowing/remembering how to set up the equation below is the hardest part of partial-fraction decomposition.

    \[\frac{-4x-3}{(x+2)^2} ~=~ \frac{A}{(x+2)}+\frac{B}{(x+2)^2}\]

    • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    \[\frac{-4x-3}{(x+2)^2} ~=~ \frac{A}{(x+2)}+\frac{B}{(x+2)^2}\]

    Multiply both sides by \((x+2)^2\).

    \[-4x-3 ~=~ A(x+2)+B\]

    Expand. Recognize the linear coefficient and the constant.

    \[-4x-3 ~=~ Ax+(2A+B)\]

    The corresponding coefficients must be equal for the equation to be true for all \(x\).

    \[\begin{align} -4 ~&=~ A \\ -3 ~&=~ 2A+B \end{align}\]

    We already have \(A\). Substitute to get \(B\). \[-3 ~=~ 2(-4)+B\]

    Multiply 2 and -4.

    \[-3 ~=~ -8+B\]

    Add \(8\) to both sides.

    \[5 ~=~ B\]

    
    
18. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. One version of partial-fraction decomposition occurs when there is a squared linear factor.

    The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restriction \(x\ne 4\)). It should be said that knowing/remembering how to set up the equation below is the hardest part of partial-fraction decomposition.

    \[\frac{8x-34}{(x-4)^2} ~=~ \frac{A}{(x-4)}+\frac{B}{(x-4)^2}\]

    • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    \[\frac{8x-34}{(x-4)^2} ~=~ \frac{A}{(x-4)}+\frac{B}{(x-4)^2}\]

    Multiply both sides by \((x-4)^2\).

    \[8x-34 ~=~ A(x-4)+B\]

    Expand. Recognize the linear coefficient and the constant.

    \[8x-34 ~=~ Ax+(-4A+B)\]

    The corresponding coefficients must be equal for the equation to be true for all \(x\).

    \[\begin{align} 8 ~&=~ A \\ -34 ~&=~ -4A+B \end{align}\]

    We already have \(A\). Substitute to get \(B\). \[-34 ~=~ -4(8)+B\]

    Multiply -4 and 8.

    \[-34 ~=~ -32+B\]

    Add \(32\) to both sides.

    \[-2 ~=~ B\]

    
    
19. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. One version of partial-fraction decomposition occurs when there is a squared linear factor.

    The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restriction \(x\ne -6\)). It should be said that knowing/remembering how to set up the equation below is the hardest part of partial-fraction decomposition.

    \[\frac{-9x-50}{(x+6)^2} ~=~ \frac{A}{(x+6)}+\frac{B}{(x+6)^2}\]

    • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    \[\frac{-9x-50}{(x+6)^2} ~=~ \frac{A}{(x+6)}+\frac{B}{(x+6)^2}\]

    Multiply both sides by \((x+6)^2\).

    \[-9x-50 ~=~ A(x+6)+B\]

    Expand. Recognize the linear coefficient and the constant.

    \[-9x-50 ~=~ Ax+(6A+B)\]

    The corresponding coefficients must be equal for the equation to be true for all \(x\).

    \[\begin{align} -9 ~&=~ A \\ -50 ~&=~ 6A+B \end{align}\]

    We already have \(A\). Substitute to get \(B\). \[-50 ~=~ 6(-9)+B\]

    Multiply 6 and -9.

    \[-50 ~=~ -54+B\]

    Add \(54\) to both sides.

    \[4 ~=~ B\]

    
    
20. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition. One version of partial-fraction decomposition occurs when there is a squared linear factor.

    The procedure can be exemplified by trying to determine constants \(A\) and \(B\) to satisfy the equation below for all values of \(x\) (with restriction \(x\ne 3\)). It should be said that knowing/remembering how to set up the equation below is the hardest part of partial-fraction decomposition.

    \[\frac{9x-31}{(x-3)^2} ~=~ \frac{A}{(x-3)}+\frac{B}{(x-3)^2}\]

    • \(A=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(B=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    \[\frac{9x-31}{(x-3)^2} ~=~ \frac{A}{(x-3)}+\frac{B}{(x-3)^2}\]

    Multiply both sides by \((x-3)^2\).

    \[9x-31 ~=~ A(x-3)+B\]

    Expand. Recognize the linear coefficient and the constant.

    \[9x-31 ~=~ Ax+(-3A+B)\]

    The corresponding coefficients must be equal for the equation to be true for all \(x\).

    \[\begin{align} 9 ~&=~ A \\ -31 ~&=~ -3A+B \end{align}\]

    We already have \(A\). Substitute to get \(B\). \[-31 ~=~ -3(9)+B\]

    Multiply -3 and 9.

    \[-31 ~=~ -27+B\]

    Add \(27\) to both sides.

    \[-4 ~=~ B\]

    
    
21. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} -5x-5y-5z&=-15 \\ -6x-4y-6z&=-4 \\ -x-2y+4z&=-10 \end{align}\]

    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    From OpenStax:

    Given a linear system of three equations, solve for three unknowns.

    • Pick any pair of equations and eliminate one variable.
    • Pick another pair of equations and eliminate the same variable.
    • You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
    • Back-substitute known variables into any one of the original equations and solve for the missing variable.

    \[x=-4\]

    \[y=7\]

    \[z=0\]

    
    
22. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} 3x+3y+4z&=-6 \\ 4x+4y+3z&=-1 \\ 5x+6y-z&=16 \end{align}\]

    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    From OpenStax:

    Given a linear system of three equations, solve for three unknowns.

    • Pick any pair of equations and eliminate one variable.
    • Pick another pair of equations and eliminate the same variable.
    • You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
    • Back-substitute known variables into any one of the original equations and solve for the missing variable.

    \[x=-1\]

    \[y=3\]

    \[z=-3\]

    
    
23. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} 6x-2y+6z&=-12 \\ x-2y+2z&=1 \\ -4x-y+6z&=-5 \end{align}\]

    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    From OpenStax:

    Given a linear system of three equations, solve for three unknowns.

    • Pick any pair of equations and eliminate one variable.
    • Pick another pair of equations and eliminate the same variable.
    • You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
    • Back-substitute known variables into any one of the original equations and solve for the missing variable.

    \[x=-1\]

    \[y=-3\]

    \[z=-2\]

    
    
24. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} x+2y+5z&=-14 \\ -3x+y-2z&=-9 \\ 5x-2y+3z&=16 \end{align}\]

    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    From OpenStax:

    Given a linear system of three equations, solve for three unknowns.

    • Pick any pair of equations and eliminate one variable.
    • Pick another pair of equations and eliminate the same variable.
    • You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
    • Back-substitute known variables into any one of the original equations and solve for the missing variable.

    \[x=7\]

    \[y=2\]

    \[z=-5\]

    
    
25. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} -3x+5y-5z&=19 \\ -4x-y-z&=17 \\ -2x+6y-5z&=18 \end{align}\]

    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    From OpenStax:

    Given a linear system of three equations, solve for three unknowns.

    • Pick any pair of equations and eliminate one variable.
    • Pick another pair of equations and eliminate the same variable.
    • You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
    • Back-substitute known variables into any one of the original equations and solve for the missing variable.

    \[x=-8\]

    \[y=7\]

    \[z=8\]

    
    
26. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} -2x+2y+z&=-4 \\ -6x-3y-6z&=-3 \\ 2x-y+2z&=3 \end{align}\]

    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    From OpenStax:

    Given a linear system of three equations, solve for three unknowns.

    • Pick any pair of equations and eliminate one variable.
    • Pick another pair of equations and eliminate the same variable.
    • You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
    • Back-substitute known variables into any one of the original equations and solve for the missing variable.

    \[x=1\]

    \[y=-1\]

    \[z=0\]

    
    
27. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} 4x-2y-3z&=-16 \\ -6x+2y+z&=10 \\ -6x-2y-5z&=-14 \end{align}\]

    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    From OpenStax:

    Given a linear system of three equations, solve for three unknowns.

    • Pick any pair of equations and eliminate one variable.
    • Pick another pair of equations and eliminate the same variable.
    • You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
    • Back-substitute known variables into any one of the original equations and solve for the missing variable.

    \[x=-1\]

    \[y=0\]

    \[z=4\]

    
    
28. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} 2x+2y-6z&=14 \\ -2x+4y-3z&=-2 \\ 6x+6y+4z&=-2 \end{align}\]

    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    From OpenStax:

    Given a linear system of three equations, solve for three unknowns.

    • Pick any pair of equations and eliminate one variable.
    • Pick another pair of equations and eliminate the same variable.
    • You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
    • Back-substitute known variables into any one of the original equations and solve for the missing variable.

    \[x=2\]

    \[y=-1\]

    \[z=-2\]

    
    
29. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} -2x-y+3z&=-16 \\ 2x-2y+2z&=6 \\ 5x-2y+4z&=17 \end{align}\]

    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    From OpenStax:

    Given a linear system of three equations, solve for three unknowns.

    • Pick any pair of equations and eliminate one variable.
    • Pick another pair of equations and eliminate the same variable.
    • You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
    • Back-substitute known variables into any one of the original equations and solve for the missing variable.

    \[x=5\]

    \[y=0\]

    \[z=-2\]

    
    
30. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} -4x+3y-4z&=-4 \\ -6x-5y-3z&=9 \\ 6x-3y+4z&=-4 \end{align}\]

    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    From OpenStax:

    Given a linear system of three equations, solve for three unknowns.

    • Pick any pair of equations and eliminate one variable.
    • Pick another pair of equations and eliminate the same variable.
    • You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
    • Back-substitute known variables into any one of the original equations and solve for the missing variable.

    \[x=-4\]

    \[y=0\]

    \[z=5\]

    
    
31. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} -w-3x-6y-2z&=25 \\ -3w+5x-y-5z&=1 \\ -3w-3x+4y-3z&=-59 \\ 4w-2x-y-2z&=-16 \end{align}\]

    • \(w=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    Don’t do this by hand. Let’s use SageMath (a beefed up Python for mathematics).

    You can run SageMath code here or here.

    var("w x y z")  #these will be our algebraic variables
    eq1 = -w-3*x-6*y-2*z==25
    eq2 = -3*w+5*x-y-5*z==1
    eq3 = -3*w-3*x+4*y-3*z==-59
    eq4 = 4*w-2*x-y-2*z==-16
    sol = solve([eq1,eq2,eq3,eq4],[w,x,y,z])
    print(sol)

    Anyway, SageMath should tell you the answer:

    \[w=-1\]

    \[x=4\]

    \[y=-8\]

    \[z=6\]

    
    
32. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} 4w+3x+y-5z&=22 \\ w+x+y-z&=12 \\ -2w-2x-5y-z&=-51 \\ -4w+x-y-z&=-26 \end{align}\]

    • \(w=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    Don’t do this by hand. Let’s use SageMath (a beefed up Python for mathematics).

    You can run SageMath code here or here.

    var("w x y z")  #these will be our algebraic variables
    eq1 = 4*w+3*x+y-5*z==22
    eq2 = w+x+y-z==12
    eq3 = -2*w-2*x-5*y-z==-51
    eq4 = -4*w+x-y-z==-26
    sol = solve([eq1,eq2,eq3,eq4],[w,x,y,z])
    print(sol)

    Anyway, SageMath should tell you the answer:

    \[w=4\]

    \[x=-1\]

    \[y=9\]

    \[z=0\]

    
    
33. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} w+3x-4y+3z&=83 \\ -3w-4x-2y-3z&=-41 \\ 6w+x+6y-6z&=-76 \\ -6w-2x-2y+z&=-3 \end{align}\]

    • \(w=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    Don’t do this by hand. Let’s use SageMath (a beefed up Python for mathematics).

    You can run SageMath code here or here.

    var("w x y z")  #these will be our algebraic variables
    eq1 = w+3*x-4*y+3*z==83
    eq2 = -3*w-4*x-2*y-3*z==-41
    eq3 = 6*w+x+6*y-6*z==-76
    eq4 = -6*w-2*x-2*y+z==-3
    sol = solve([eq1,eq2,eq3,eq4],[w,x,y,z])
    print(sol)

    Anyway, SageMath should tell you the answer:

    \[w=2\]

    \[x=8\]

    \[y=-9\]

    \[z=7\]

    
    
34. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} 2w-4x+y+2z&=-3 \\ 2w-5x+4y+5z&=28 \\ -5w+x-6y+z&=-51 \\ -3w+x-6y-6z&=-70 \end{align}\]

    • \(w=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    Don’t do this by hand. Let’s use SageMath (a beefed up Python for mathematics).

    You can run SageMath code here or here.

    var("w x y z")  #these will be our algebraic variables
    eq1 = 2*w-4*x+y+2*z==-3
    eq2 = 2*w-5*x+4*y+5*z==28
    eq3 = -5*w+x-6*y+z==-51
    eq4 = -3*w+x-6*y-6*z==-70
    sol = solve([eq1,eq2,eq3,eq4],[w,x,y,z])
    print(sol)

    Anyway, SageMath should tell you the answer:

    \[w=1\]

    \[x=5\]

    \[y=9\]

    \[z=3\]

    
    
35. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} 4w+3x+6y-6z&=77 \\ -w-x-2y+5z&=-41 \\ 4w-6x+4y-4z&=38 \\ 2w-2x+2y+3z&=-8 \end{align}\]

    • \(w=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    Don’t do this by hand. Let’s use SageMath (a beefed up Python for mathematics).

    You can run SageMath code here or here.

    var("w x y z")  #these will be our algebraic variables
    eq1 = 4*w+3*x+6*y-6*z==77
    eq2 = -w-x-2*y+5*z==-41
    eq3 = 4*w-6*x+4*y-4*z==38
    eq4 = 2*w-2*x+2*y+3*z==-8
    sol = solve([eq1,eq2,eq3,eq4],[w,x,y,z])
    print(sol)

    Anyway, SageMath should tell you the answer:

    \[w=8\]

    \[x=3\]

    \[y=0\]

    \[z=-6\]

    
    
36. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} 6w+6x-2y-6z&=-28 \\ 3w-3x+3y-4z&=64 \\ -5w+4x+5y-4z&=23 \\ w+x-y+3z&=-26 \end{align}\]

    • \(w=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    Don’t do this by hand. Let’s use SageMath (a beefed up Python for mathematics).

    You can run SageMath code here or here.

    var("w x y z")  #these will be our algebraic variables
    eq1 = 6*w+6*x-2*y-6*z==-28
    eq2 = 3*w-3*x+3*y-4*z==64
    eq3 = -5*w+4*x+5*y-4*z==23
    eq4 = w+x-y+3*z==-26
    sol = solve([eq1,eq2,eq3,eq4],[w,x,y,z])
    print(sol)

    Anyway, SageMath should tell you the answer:

    \[w=1\]

    \[x=-7\]

    \[y=8\]

    \[z=-4\]

    
    
37. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} 6w+3x+2y+z&=4 \\ w+6x-4y-6z&=-10 \\ -2w-x-y+4z&=-3 \\ -6w-x+4y+6z&=30 \end{align}\]

    • \(w=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    Don’t do this by hand. Let’s use SageMath (a beefed up Python for mathematics).

    You can run SageMath code here or here.

    var("w x y z")  #these will be our algebraic variables
    eq1 = 6*w+3*x+2*y+z==4
    eq2 = w+6*x-4*y-6*z==-10
    eq3 = -2*w-x-y+4*z==-3
    eq4 = -6*w-x+4*y+6*z==30
    sol = solve([eq1,eq2,eq3,eq4],[w,x,y,z])
    print(sol)

    Anyway, SageMath should tell you the answer:

    \[w=-2\]

    \[x=2\]

    \[y=5\]

    \[z=0\]

    
    
38. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} -w+4x+6y-2z&=-67 \\ 3w+2x+2y+3z&=-49 \\ 3w-4x-4y+4z&=37 \\ -6w+5x+2y+4z&=16 \end{align}\]

    • \(w=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    Don’t do this by hand. Let’s use SageMath (a beefed up Python for mathematics).

    You can run SageMath code here or here.

    var("w x y z")  #these will be our algebraic variables
    eq1 = -w+4*x+6*y-2*z==-67
    eq2 = 3*w+2*x+2*y+3*z==-49
    eq3 = 3*w-4*x-4*y+4*z==37
    eq4 = -6*w+5*x+2*y+4*z==16
    sol = solve([eq1,eq2,eq3,eq4],[w,x,y,z])
    print(sol)

    Anyway, SageMath should tell you the answer:

    \[w=-9\]

    \[x=-6\]

    \[y=-8\]

    \[z=2\]

    
    
39. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} w-5x-3y+3z&=-7 \\ 4w-4x-y+z&=-41 \\ 4w-3x-2y-5z&=-30 \\ -4w+x-3y-z&=57 \end{align}\]

    • \(w=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    Don’t do this by hand. Let’s use SageMath (a beefed up Python for mathematics).

    You can run SageMath code here or here.

    var("w x y z")  #these will be our algebraic variables
    eq1 = w-5*x-3*y+3*z==-7
    eq2 = 4*w-4*x-y+z==-41
    eq3 = 4*w-3*x-2*y-5*z==-30
    eq4 = -4*w+x-3*y-z==57
    sol = solve([eq1,eq2,eq3,eq4],[w,x,y,z])
    print(sol)

    Anyway, SageMath should tell you the answer:

    \[w=-8\]

    \[x=4\]

    \[y=-7\]

    \[z=0\]

    
    
40. 
    

    Question

    prevnext

    Solve the system of linear equations.

    \[\begin{align} -4w+x+2y+4z&=-16 \\ -4w-4x-3y+3z&=5 \\ -4w+5x+6y+4z&=-32 \\ -2w+6x-2y+z&=-63 \end{align}\]

    • \(w=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(x=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(y=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(z=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    Don’t do this by hand. Let’s use SageMath (a beefed up Python for mathematics).

    You can run SageMath code here or here.

    var("w x y z")  #these will be our algebraic variables
    eq1 = -4*w+x+2*y+4*z==-16
    eq2 = -4*w-4*x-3*y+3*z==5
    eq3 = -4*w+5*x+6*y+4*z==-32
    eq4 = -2*w+6*x-2*y+z==-63
    sol = solve([eq1,eq2,eq3,eq4],[w,x,y,z])
    print(sol)

    Anyway, SageMath should tell you the answer:

    \[w=3\]

    \[x=-8\]

    \[y=4\]

    \[z=-1\]

    
    
41. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition.

    The procedure can be exemplified by trying to determine constants \(A\), \(B\), and \(C\) to satisfy the equation below for all values of \(x\).

    \[\frac{-13x^2-14x+3}{(x+3)(x-5)(x+2)} = \frac{A}{x+3}+\frac{B}{x-5}+\frac{C}{x+2}\]

    • \(A=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(B=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(C=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    We start with the given equivalence: \[\frac{-13x^2-14x+3}{(x+3)(x-5)(x+2)} = \frac{A}{x+3}+\frac{B}{x-5}+\frac{C}{x+2}\]

    Multiply both sides by \((x+3)(x-5)(x+2)\).

    \[-13x^2-14x+3 ~~=~~ A(x-5)(x+2)+B(x+3)(x+2)+C(x+3)(x-5)\]

    Expand (by FOIL and distributing).

    \[-13x^2-14x+3 ~~=~~ Ax^2-3Ax-10A+Bx^2+5Bx+6B+Cx^2-2Cx-15C\]

    Gather similar terms (with same degree of \(x\)) to identify the coefficients.

    \[(-13)x^2+(-14)x+(3) ~~=~~ (A+B+C)x^2+(-3A+5B-2C)x+(-10A+6B-15C)\] Corresponding coefficients must be equal for the above equivalence to hold for all \(x\).

    \[\begin{align} -13&=A+B+C\\ -14&=-3A+5B-2C\\ 3&=-10A+6B-15C \end{align}\]

    Solve the 3x3 system. You can use elimination, or you can use SageMath.

    var("A B C")
    eq1 = -13==A+B+C
    eq2 = -14==-3*A+5*B-2*C
    eq3 = 3==-10*A+6*B-15*C
    solve([eq1,eq2,eq3],[A,B,C])

    \[\begin{align} A&=-9\\ B&=-7\\ C&=3 \end{align}\]

    And, really, SageMath can do the entire partial-fraction decomposition by using the partial_fraction() method available to expression objects.

    var("x")
    nmrtr = -13*x^2-14*x+3
    dnmntr = (x+3)*(x-5)*(x+2)
    exp = nmrtr/dnmntr
    exp.partial_fraction()

    
    
42. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition.

    The procedure can be exemplified by trying to determine constants \(A\), \(B\), and \(C\) to satisfy the equation below for all values of \(x\).

    \[\frac{-4x^2-9x+13}{(x+2)(x+5)(x+3)} = \frac{A}{x+2}+\frac{B}{x+5}+\frac{C}{x+3}\]

    • \(A=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(B=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(C=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    We start with the given equivalence: \[\frac{-4x^2-9x+13}{(x+2)(x+5)(x+3)} = \frac{A}{x+2}+\frac{B}{x+5}+\frac{C}{x+3}\]

    Multiply both sides by \((x+2)(x+5)(x+3)\).

    \[-4x^2-9x+13 ~~=~~ A(x+5)(x+3)+B(x+2)(x+3)+C(x+2)(x+5)\]

    Expand (by FOIL and distributing).

    \[-4x^2-9x+13 ~~=~~ Ax^2+8Ax+15A+Bx^2+5Bx+6B+Cx^2+7Cx+10C\]

    Gather similar terms (with same degree of \(x\)) to identify the coefficients.

    \[(-4)x^2+(-9)x+(13) ~~=~~ (A+B+C)x^2+(8A+5B+7C)x+(15A+6B+10C)\] Corresponding coefficients must be equal for the above equivalence to hold for all \(x\).

    \[\begin{align} -4&=A+B+C\\ -9&=8A+5B+7C\\ 13&=15A+6B+10C \end{align}\]

    Solve the 3x3 system. You can use elimination, or you can use SageMath.

    var("A B C")
    eq1 = -4==A+B+C
    eq2 = -9==8*A+5*B+7*C
    eq3 = 13==15*A+6*B+10*C
    solve([eq1,eq2,eq3],[A,B,C])

    \[\begin{align} A&=5\\ B&=-7\\ C&=-2 \end{align}\]

    And, really, SageMath can do the entire partial-fraction decomposition by using the partial_fraction() method available to expression objects.

    var("x")
    nmrtr = -4*x^2-9*x+13
    dnmntr = (x+2)*(x+5)*(x+3)
    exp = nmrtr/dnmntr
    exp.partial_fraction()

    
    
43. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition.

    The procedure can be exemplified by trying to determine constants \(A\), \(B\), and \(C\) to satisfy the equation below for all values of \(x\).

    \[\frac{-5x^2-14x+12}{(x+6)(x+3)(x+2)} = \frac{A}{x+6}+\frac{B}{x+3}+\frac{C}{x+2}\]

    • \(A=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(B=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(C=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    We start with the given equivalence: \[\frac{-5x^2-14x+12}{(x+6)(x+3)(x+2)} = \frac{A}{x+6}+\frac{B}{x+3}+\frac{C}{x+2}\]

    Multiply both sides by \((x+6)(x+3)(x+2)\).

    \[-5x^2-14x+12 ~~=~~ A(x+3)(x+2)+B(x+6)(x+2)+C(x+6)(x+3)\]

    Expand (by FOIL and distributing).

    \[-5x^2-14x+12 ~~=~~ Ax^2+5Ax+6A+Bx^2+8Bx+12B+Cx^2+9Cx+18C\]

    Gather similar terms (with same degree of \(x\)) to identify the coefficients.

    \[(-5)x^2+(-14)x+(12) ~~=~~ (A+B+C)x^2+(5A+8B+9C)x+(6A+12B+18C)\] Corresponding coefficients must be equal for the above equivalence to hold for all \(x\).

    \[\begin{align} -5&=A+B+C\\ -14&=5A+8B+9C\\ 12&=6A+12B+18C \end{align}\]

    Solve the 3x3 system. You can use elimination, or you can use SageMath.

    var("A B C")
    eq1 = -5==A+B+C
    eq2 = -14==5*A+8*B+9*C
    eq3 = 12==6*A+12*B+18*C
    solve([eq1,eq2,eq3],[A,B,C])

    \[\begin{align} A&=-7\\ B&=-3\\ C&=5 \end{align}\]

    And, really, SageMath can do the entire partial-fraction decomposition by using the partial_fraction() method available to expression objects.

    var("x")
    nmrtr = -5*x^2-14*x+12
    dnmntr = (x+6)*(x+3)*(x+2)
    exp = nmrtr/dnmntr
    exp.partial_fraction()

    
    
44. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition.

    The procedure can be exemplified by trying to determine constants \(A\), \(B\), and \(C\) to satisfy the equation below for all values of \(x\).

    \[\frac{-7x^2-9x+18}{(x-2)(x-3)(x+5)} = \frac{A}{x-2}+\frac{B}{x-3}+\frac{C}{x+5}\]

    • \(A=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(B=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(C=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    We start with the given equivalence: \[\frac{-7x^2-9x+18}{(x-2)(x-3)(x+5)} = \frac{A}{x-2}+\frac{B}{x-3}+\frac{C}{x+5}\]

    Multiply both sides by \((x-2)(x-3)(x+5)\).

    \[-7x^2-9x+18 ~~=~~ A(x-3)(x+5)+B(x-2)(x+5)+C(x-2)(x-3)\]

    Expand (by FOIL and distributing).

    \[-7x^2-9x+18 ~~=~~ Ax^2+2Ax-15A+Bx^2+3Bx-10B+Cx^2-5Cx+6C\]

    Gather similar terms (with same degree of \(x\)) to identify the coefficients.

    \[(-7)x^2+(-9)x+(18) ~~=~~ (A+B+C)x^2+(2A+3B-5C)x+(-15A-10B+6C)\] Corresponding coefficients must be equal for the above equivalence to hold for all \(x\).

    \[\begin{align} -7&=A+B+C\\ -9&=2A+3B-5C\\ 18&=-15A-10B+6C \end{align}\]

    Solve the 3x3 system. You can use elimination, or you can use SageMath.

    var("A B C")
    eq1 = -7==A+B+C
    eq2 = -9==2*A+3*B-5*C
    eq3 = 18==-15*A-10*B+6*C
    solve([eq1,eq2,eq3],[A,B,C])

    \[\begin{align} A&=4\\ B&=-9\\ C&=-2 \end{align}\]

    And, really, SageMath can do the entire partial-fraction decomposition by using the partial_fraction() method available to expression objects.

    var("x")
    nmrtr = -7*x^2-9*x+18
    dnmntr = (x-2)*(x-3)*(x+5)
    exp = nmrtr/dnmntr
    exp.partial_fraction()

    
    
45. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition.

    The procedure can be exemplified by trying to determine constants \(A\), \(B\), and \(C\) to satisfy the equation below for all values of \(x\).

    \[\frac{-4x^2+9x+19}{(x-2)(x-5)(x-3)} = \frac{A}{x-2}+\frac{B}{x-5}+\frac{C}{x-3}\]

    • \(A=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(B=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(C=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    We start with the given equivalence: \[\frac{-4x^2+9x+19}{(x-2)(x-5)(x-3)} = \frac{A}{x-2}+\frac{B}{x-5}+\frac{C}{x-3}\]

    Multiply both sides by \((x-2)(x-5)(x-3)\).

    \[-4x^2+9x+19 ~~=~~ A(x-5)(x-3)+B(x-2)(x-3)+C(x-2)(x-5)\]

    Expand (by FOIL and distributing).

    \[-4x^2+9x+19 ~~=~~ Ax^2-8Ax+15A+Bx^2-5Bx+6B+Cx^2-7Cx+10C\]

    Gather similar terms (with same degree of \(x\)) to identify the coefficients.

    \[(-4)x^2+(9)x+(19) ~~=~~ (A+B+C)x^2+(-8A-5B-7C)x+(15A+6B+10C)\] Corresponding coefficients must be equal for the above equivalence to hold for all \(x\).

    \[\begin{align} -4&=A+B+C\\ 9&=-8A-5B-7C\\ 19&=15A+6B+10C \end{align}\]

    Solve the 3x3 system. You can use elimination, or you can use SageMath.

    var("A B C")
    eq1 = -4==A+B+C
    eq2 = 9==-8*A-5*B-7*C
    eq3 = 19==15*A+6*B+10*C
    solve([eq1,eq2,eq3],[A,B,C])

    \[\begin{align} A&=7\\ B&=-6\\ C&=-5 \end{align}\]

    And, really, SageMath can do the entire partial-fraction decomposition by using the partial_fraction() method available to expression objects.

    var("x")
    nmrtr = -4*x^2+9*x+19
    dnmntr = (x-2)*(x-5)*(x-3)
    exp = nmrtr/dnmntr
    exp.partial_fraction()

    
    
46. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition.

    The procedure can be exemplified by trying to determine constants \(A\), \(B\), and \(C\) to satisfy the equation below for all values of \(x\).

    \[\frac{-11x^2+8x+12}{(x+6)(x-2)(x-3)} = \frac{A}{x+6}+\frac{B}{x-2}+\frac{C}{x-3}\]

    • \(A=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(B=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(C=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    We start with the given equivalence: \[\frac{-11x^2+8x+12}{(x+6)(x-2)(x-3)} = \frac{A}{x+6}+\frac{B}{x-2}+\frac{C}{x-3}\]

    Multiply both sides by \((x+6)(x-2)(x-3)\).

    \[-11x^2+8x+12 ~~=~~ A(x-2)(x-3)+B(x+6)(x-3)+C(x+6)(x-2)\]

    Expand (by FOIL and distributing).

    \[-11x^2+8x+12 ~~=~~ Ax^2-5Ax+6A+Bx^2+3Bx-18B+Cx^2+4Cx-12C\]

    Gather similar terms (with same degree of \(x\)) to identify the coefficients.

    \[(-11)x^2+(8)x+(12) ~~=~~ (A+B+C)x^2+(-5A+3B+4C)x+(6A-18B-12C)\] Corresponding coefficients must be equal for the above equivalence to hold for all \(x\).

    \[\begin{align} -11&=A+B+C\\ 8&=-5A+3B+4C\\ 12&=6A-18B-12C \end{align}\]

    Solve the 3x3 system. You can use elimination, or you can use SageMath.

    var("A B C")
    eq1 = -11==A+B+C
    eq2 = 8==-5*A+3*B+4*C
    eq3 = 12==6*A-18*B-12*C
    solve([eq1,eq2,eq3],[A,B,C])

    \[\begin{align} A&=-6\\ B&=2\\ C&=-7 \end{align}\]

    And, really, SageMath can do the entire partial-fraction decomposition by using the partial_fraction() method available to expression objects.

    var("x")
    nmrtr = -11*x^2+8*x+12
    dnmntr = (x+6)*(x-2)*(x-3)
    exp = nmrtr/dnmntr
    exp.partial_fraction()

    
    
47. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition.

    The procedure can be exemplified by trying to determine constants \(A\), \(B\), and \(C\) to satisfy the equation below for all values of \(x\).

    \[\frac{-10x^2-7x+5}{(x-5)(x+3)(x+2)} = \frac{A}{x-5}+\frac{B}{x+3}+\frac{C}{x+2}\]

    • \(A=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(B=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(C=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    We start with the given equivalence: \[\frac{-10x^2-7x+5}{(x-5)(x+3)(x+2)} = \frac{A}{x-5}+\frac{B}{x+3}+\frac{C}{x+2}\]

    Multiply both sides by \((x-5)(x+3)(x+2)\).

    \[-10x^2-7x+5 ~~=~~ A(x+3)(x+2)+B(x-5)(x+2)+C(x-5)(x+3)\]

    Expand (by FOIL and distributing).

    \[-10x^2-7x+5 ~~=~~ Ax^2+5Ax+6A+Bx^2-3Bx-10B+Cx^2-2Cx-15C\]

    Gather similar terms (with same degree of \(x\)) to identify the coefficients.

    \[(-10)x^2+(-7)x+(5) ~~=~~ (A+B+C)x^2+(5A-3B-2C)x+(6A-10B-15C)\] Corresponding coefficients must be equal for the above equivalence to hold for all \(x\).

    \[\begin{align} -10&=A+B+C\\ -7&=5A-3B-2C\\ 5&=6A-10B-15C \end{align}\]

    Solve the 3x3 system. You can use elimination, or you can use SageMath.

    var("A B C")
    eq1 = -10==A+B+C
    eq2 = -7==5*A-3*B-2*C
    eq3 = 5==6*A-10*B-15*C
    solve([eq1,eq2,eq3],[A,B,C])

    \[\begin{align} A&=-5\\ B&=-8\\ C&=3 \end{align}\]

    And, really, SageMath can do the entire partial-fraction decomposition by using the partial_fraction() method available to expression objects.

    var("x")
    nmrtr = -10*x^2-7*x+5
    dnmntr = (x-5)*(x+3)*(x+2)
    exp = nmrtr/dnmntr
    exp.partial_fraction()

    
    
48. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition.

    The procedure can be exemplified by trying to determine constants \(A\), \(B\), and \(C\) to satisfy the equation below for all values of \(x\).

    \[\frac{3x^2+11x+16}{(x+2)(x+5)(x+3)} = \frac{A}{x+2}+\frac{B}{x+5}+\frac{C}{x+3}\]

    • \(A=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(B=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(C=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    We start with the given equivalence: \[\frac{3x^2+11x+16}{(x+2)(x+5)(x+3)} = \frac{A}{x+2}+\frac{B}{x+5}+\frac{C}{x+3}\]

    Multiply both sides by \((x+2)(x+5)(x+3)\).

    \[3x^2+11x+16 ~~=~~ A(x+5)(x+3)+B(x+2)(x+3)+C(x+2)(x+5)\]

    Expand (by FOIL and distributing).

    \[3x^2+11x+16 ~~=~~ Ax^2+8Ax+15A+Bx^2+5Bx+6B+Cx^2+7Cx+10C\]

    Gather similar terms (with same degree of \(x\)) to identify the coefficients.

    \[(3)x^2+(11)x+(16) ~~=~~ (A+B+C)x^2+(8A+5B+7C)x+(15A+6B+10C)\] Corresponding coefficients must be equal for the above equivalence to hold for all \(x\).

    \[\begin{align} 3&=A+B+C\\ 11&=8A+5B+7C\\ 16&=15A+6B+10C \end{align}\]

    Solve the 3x3 system. You can use elimination, or you can use SageMath.

    var("A B C")
    eq1 = 3==A+B+C
    eq2 = 11==8*A+5*B+7*C
    eq3 = 16==15*A+6*B+10*C
    solve([eq1,eq2,eq3],[A,B,C])

    \[\begin{align} A&=2\\ B&=6\\ C&=-5 \end{align}\]

    And, really, SageMath can do the entire partial-fraction decomposition by using the partial_fraction() method available to expression objects.

    var("x")
    nmrtr = 3*x^2+11*x+16
    dnmntr = (x+2)*(x+5)*(x+3)
    exp = nmrtr/dnmntr
    exp.partial_fraction()

    
    
49. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition.

    The procedure can be exemplified by trying to determine constants \(A\), \(B\), and \(C\) to satisfy the equation below for all values of \(x\).

    \[\frac{-11x^2-11x+10}{(x-4)(x+2)(x+3)} = \frac{A}{x-4}+\frac{B}{x+2}+\frac{C}{x+3}\]

    • \(A=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(B=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(C=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    We start with the given equivalence: \[\frac{-11x^2-11x+10}{(x-4)(x+2)(x+3)} = \frac{A}{x-4}+\frac{B}{x+2}+\frac{C}{x+3}\]

    Multiply both sides by \((x-4)(x+2)(x+3)\).

    \[-11x^2-11x+10 ~~=~~ A(x+2)(x+3)+B(x-4)(x+3)+C(x-4)(x+2)\]

    Expand (by FOIL and distributing).

    \[-11x^2-11x+10 ~~=~~ Ax^2+5Ax+6A+Bx^2-Bx-12B+Cx^2-2Cx-8C\]

    Gather similar terms (with same degree of \(x\)) to identify the coefficients.

    \[(-11)x^2+(-11)x+(10) ~~=~~ (A+B+C)x^2+(5A-B-2C)x+(6A-12B-8C)\] Corresponding coefficients must be equal for the above equivalence to hold for all \(x\).

    \[\begin{align} -11&=A+B+C\\ -11&=5A-B-2C\\ 10&=6A-12B-8C \end{align}\]

    Solve the 3x3 system. You can use elimination, or you can use SageMath.

    var("A B C")
    eq1 = -11==A+B+C
    eq2 = -11==5*A-B-2*C
    eq3 = 10==6*A-12*B-8*C
    solve([eq1,eq2,eq3],[A,B,C])

    \[\begin{align} A&=-5\\ B&=2\\ C&=-8 \end{align}\]

    And, really, SageMath can do the entire partial-fraction decomposition by using the partial_fraction() method available to expression objects.

    var("x")
    nmrtr = -11*x^2-11*x+10
    dnmntr = (x-4)*(x+2)*(x+3)
    exp = nmrtr/dnmntr
    exp.partial_fraction()

    
    
50. 
    

    Question

    prevnext

    In Calculus, when integrating rational expressions, it is often helpful to perform partial-fraction decomposition.

    The procedure can be exemplified by trying to determine constants \(A\), \(B\), and \(C\) to satisfy the equation below for all values of \(x\).

    \[\frac{2x^2-3x+18}{(x-6)(x-3)(x-2)} = \frac{A}{x-6}+\frac{B}{x-3}+\frac{C}{x-2}\]

    • \(A=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(B=\) -------- -9-8-7-6-5-4-3-2-10123456789
    • \(C=\) -------- -9-8-7-6-5-4-3-2-10123456789
    
    
    
    

    Solution

    Show solution

    We start with the given equivalence: \[\frac{2x^2-3x+18}{(x-6)(x-3)(x-2)} = \frac{A}{x-6}+\frac{B}{x-3}+\frac{C}{x-2}\]

    Multiply both sides by \((x-6)(x-3)(x-2)\).

    \[2x^2-3x+18 ~~=~~ A(x-3)(x-2)+B(x-6)(x-2)+C(x-6)(x-3)\]

    Expand (by FOIL and distributing).

    \[2x^2-3x+18 ~~=~~ Ax^2-5Ax+6A+Bx^2-8Bx+12B+Cx^2-9Cx+18C\]

    Gather similar terms (with same degree of \(x\)) to identify the coefficients.

    \[(2)x^2+(-3)x+(18) ~~=~~ (A+B+C)x^2+(-5A-8B-9C)x+(6A+12B+18C)\] Corresponding coefficients must be equal for the above equivalence to hold for all \(x\).

    \[\begin{align} 2&=A+B+C\\ -3&=-5A-8B-9C\\ 18&=6A+12B+18C \end{align}\]

    Solve the 3x3 system. You can use elimination, or you can use SageMath.

    var("A B C")
    eq1 = 2==A+B+C
    eq2 = -3==-5*A-8*B-9*C
    eq3 = 18==6*A+12*B+18*C
    solve([eq1,eq2,eq3],[A,B,C])

    \[\begin{align} A&=6\\ B&=-9\\ C&=5 \end{align}\]

    And, really, SageMath can do the entire partial-fraction decomposition by using the partial_fraction() method available to expression objects.

    var("x")
    nmrtr = 2*x^2-3*x+18
    dnmntr = (x-6)*(x-3)*(x-2)
    exp = nmrtr/dnmntr
    exp.partial_fraction()

    
    
51. 
    

    Question

    prevnext

    A matrix is a rectangular array of numbers. The dimensions of a matrix represent the number of rows and the number of columns (in that order!). Find the dimensions of the matrix below.

    \[\begin{bmatrix} -7 & 2 & 2\\ -8 & -8 & -2 \end{bmatrix}\]

    Dimensions: -------- 12345678910 \(\times\)-------- 12345678910

    
    
    
    

    Solution

    Show solution

    There are 2 rows and 3 columns. We say, “It is a 2 by 3 matrix.” We write the dimensions as 2 \(\times\) 3.

    
    
52. 
    

    Question

    prevnext

    A matrix is a rectangular array of numbers. The dimensions of a matrix represent the number of rows and the number of columns (in that order!). Find the dimensions of the matrix below.

    \[\begin{bmatrix} 6 & -8 & 1\\ -3 & 5 & 9\\ -2 & 9 & 8\\ 1 & -5 & 3\\ 0 & -4 & 3\\ -3 & -8 & -3\\ -2 & 5 & 1 \end{bmatrix}\]

    Dimensions: -------- 12345678910 \(\times\)-------- 12345678910

    
    
    
    

    Solution

    Show solution

    There are 7 rows and 3 columns. We say, “It is a 7 by 3 matrix.” We write the dimensions as 7 \(\times\) 3.

    
    
53. 
    

    Question

    prevnext

    A matrix is a rectangular array of numbers. The dimensions of a matrix represent the number of rows and the number of columns (in that order!). Find the dimensions of the matrix below.

    \[\begin{bmatrix} 4 & -7 & -7 & 0 & -7 & -2 & -7\\ -4 & 1 & -6 & 5 & -2 & -9 & -9\\ 7 & -8 & 9 & 5 & 9 & -6 & -1\\ -2 & 7 & 3 & 3 & 5 & -4 & 8 \end{bmatrix}\]

    Dimensions: -------- 12345678910 \(\times\)-------- 12345678910

    
    
    
    

    Solution

    Show solution

    There are 4 rows and 7 columns. We say, “It is a 4 by 7 matrix.” We write the dimensions as 4 \(\times\) 7.

    
    
54. 
    

    Question

    prevnext

    A matrix is a rectangular array of numbers. The dimensions of a matrix represent the number of rows and the number of columns (in that order!). Find the dimensions of the matrix below.

    \[\begin{bmatrix} 2 & 8 & -4 & -1 & 8 & -8 & -2\\ 3 & 1 & 3 & 7 & -4 & -9 & 0 \end{bmatrix}\]

    Dimensions: -------- 12345678910 \(\times\)-------- 12345678910

    
    
    
    

    Solution

    Show solution

    There are 2 rows and 7 columns. We say, “It is a 2 by 7 matrix.” We write the dimensions as 2 \(\times\) 7.

    
    
55. 
    

    Question

    prevnext

    A matrix is a rectangular array of numbers. The dimensions of a matrix represent the number of rows and the number of columns (in that order!). Find the dimensions of the matrix below.

    \[\begin{bmatrix} 6 & -5 & -3 & -1 & 0 & 6 & 9\\ -5 & 3 & 2 & 1 & 1 & 2 & -5\\ 9 & -7 & 1 & 4 & -2 & 9 & 3\\ -7 & 9 & 1 & -2 & 7 & 5 & -3 \end{bmatrix}\]

    Dimensions: -------- 12345678910 \(\times\)-------- 12345678910

    
    
    
    

    Solution

    Show solution

    There are 4 rows and 7 columns. We say, “It is a 4 by 7 matrix.” We write the dimensions as 4 \(\times\) 7.

    
    
56. 
    

    Question

    prevnext

    A matrix is a rectangular array of numbers. The dimensions of a matrix represent the number of rows and the number of columns (in that order!). Find the dimensions of the matrix below.

    \[\begin{bmatrix} 7 & 4 & -8 & 0 & 8 & 9 & -6\\ 3 & -4 & 8 & 0 & 2 & -5 & -3\\ -1 & -7 & 2 & 8 & -9 & 2 & 5\\ -8 & 2 & -6 & -8 & -6 & 5 & 6 \end{bmatrix}\]

    Dimensions: -------- 12345678910 \(\times\)-------- 12345678910

    
    
    
    

    Solution

    Show solution

    There are 4 rows and 7 columns. We say, “It is a 4 by 7 matrix.” We write the dimensions as 4 \(\times\) 7.

    
    
57. 
    

    Question

    prevnext

    A matrix is a rectangular array of numbers. The dimensions of a matrix represent the number of rows and the number of columns (in that order!). Find the dimensions of the matrix below.

    \[\begin{bmatrix} -1 & -1 & -2\\ 4 & -3 & 9\\ -3 & 8 & -5\\ 0 & -4 & -2 \end{bmatrix}\]

    Dimensions: -------- 12345678910 \(\times\)-------- 12345678910

    
    
    
    

    Solution

    Show solution

    There are 4 rows and 3 columns. We say, “It is a 4 by 3 matrix.” We write the dimensions as 4 \(\times\) 3.

    
    
58. 
    

    Question

    prevnext

    A matrix is a rectangular array of numbers. The dimensions of a matrix represent the number of rows and the number of columns (in that order!). Find the dimensions of the matrix below.

    \[\begin{bmatrix} 3 & 2 & 2 & -5 & 7 & -3\\ 7 & -5 & -5 & 5 & 2 & 3\\ -8 & 3 & 4 & -7 & -7 & 8 \end{bmatrix}\]

    Dimensions: -------- 12345678910 \(\times\)-------- 12345678910

    
    
    
    

    Solution

    Show solution

    There are 3 rows and 6 columns. We say, “It is a 3 by 6 matrix.” We write the dimensions as 3 \(\times\) 6.

    
    
59. 
    

    Question

    prevnext

    A matrix is a rectangular array of numbers. The dimensions of a matrix represent the number of rows and the number of columns (in that order!). Find the dimensions of the matrix below.

    \[\begin{bmatrix} 9 & 3 & 8 & -1 & -6\\ -9 & 2 & 5 & 4 & -4 \end{bmatrix}\]

    Dimensions: -------- 12345678910 \(\times\)-------- 12345678910

    
    
    
    

    Solution

    Show solution

    There are 2 rows and 5 columns. We say, “It is a 2 by 5 matrix.” We write the dimensions as 2 \(\times\) 5.

    
    
60. 
    

    Question

    prevnext

    A matrix is a rectangular array of numbers. The dimensions of a matrix represent the number of rows and the number of columns (in that order!). Find the dimensions of the matrix below.

    \[\begin{bmatrix} -8 & 4 & -9 & -5\\ 0 & 3 & 8 & 3\\ -8 & 0 & -4 & 2 \end{bmatrix}\]

    Dimensions: -------- 12345678910 \(\times\)-------- 12345678910

    
    
    
    

    Solution

    Show solution

    There are 3 rows and 4 columns. We say, “It is a 3 by 4 matrix.” We write the dimensions as 3 \(\times\) 4.

    
    
61. 
    

    Question

    prevnext

    If the dimensions of matrix \(A\) are \(m\times n\) and the dimensions of matrix \(B\) are \(p\times q\), then the matrix-multiplication \(AB\) is only possible if \(n=p\).

    \[A=\begin{bmatrix} -8 & -2\\ -2 & -2\\ -6 & 2 \end{bmatrix}\] \[B=\begin{bmatrix} 2 & 3 & 5 & 3\\ -6 & 7 & 4 & -3\\ -8 & 3 & -7 & -6\\ -9 & -5 & -1 & 0\\ 0 & -1 & 5 & 3\\ 5 & 7 & 2 & -6 \end{bmatrix}\]

    • \(\mathrm{dim}(A)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • \(\mathrm{dim}(B)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • Is the product \(AB\) defined? -------- Yes, the matrices can be multiplied.No, the matrices cannot be multiplied.
    
    
    
    

    Solution

    Show solution
    • The dimensions of \(A\) are \(3\times2\).
    • The dimensions of \(B\) are \(6\times4\).
    • The number of columns of \(A\) is NOT equal to the number of rows of \(B\), so we can NOT find the product \(AB\).

    
    
62. 
    

    Question

    prevnext

    If the dimensions of matrix \(A\) are \(m\times n\) and the dimensions of matrix \(B\) are \(p\times q\), then the matrix-multiplication \(AB\) is only possible if \(n=p\).

    \[A=\begin{bmatrix} 7 & 3 & -4 & -1 & 6 & 8\\ 6 & -4 & 4 & -2 & -1 & 6 \end{bmatrix}\] \[B=\begin{bmatrix} -4 & 3 & -3 & 0\\ -6 & 4 & 9 & 2\\ -8 & -3 & 2 & -9 \end{bmatrix}\]

    • \(\mathrm{dim}(A)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • \(\mathrm{dim}(B)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • Is the product \(AB\) defined? -------- Yes, the matrices can be multiplied.No, the matrices cannot be multiplied.
    
    
    
    

    Solution

    Show solution
    • The dimensions of \(A\) are \(2\times6\).
    • The dimensions of \(B\) are \(3\times4\).
    • The number of columns of \(A\) is NOT equal to the number of rows of \(B\), so we can NOT find the product \(AB\).

    
    
63. 
    

    Question

    prevnext

    If the dimensions of matrix \(A\) are \(m\times n\) and the dimensions of matrix \(B\) are \(p\times q\), then the matrix-multiplication \(AB\) is only possible if \(n=p\).

    \[A=\begin{bmatrix} 7 & -5 & 6 & 2\\ 8 & 0 & 7 & 0 \end{bmatrix}\] \[B=\begin{bmatrix} 8 & 1 & 2\\ 9 & 7 & -6\\ 3 & 9 & 3\\ -4 & 9 & 3 \end{bmatrix}\]

    • \(\mathrm{dim}(A)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • \(\mathrm{dim}(B)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • Is the product \(AB\) defined? -------- Yes, the matrices can be multiplied.No, the matrices cannot be multiplied.
    
    
    
    

    Solution

    Show solution
    • The dimensions of \(A\) are \(2\times4\).
    • The dimensions of \(B\) are \(4\times3\).
    • The number of columns of \(A\) is equal to the number of rows of \(B\), so we can find the product \(AB\).

    The dimensions of the product \(C=AB\) would be \(2\times3\).

    
    
64. 
    

    Question

    prevnext

    If the dimensions of matrix \(A\) are \(m\times n\) and the dimensions of matrix \(B\) are \(p\times q\), then the matrix-multiplication \(AB\) is only possible if \(n=p\).

    \[A=\begin{bmatrix} -4 & -8\\ 2 & 5\\ -7 & -6 \end{bmatrix}\] \[B=\begin{bmatrix} 0 & -8 & -3 & -1 & -2\\ 6 & 6 & 0 & -6 & -6 \end{bmatrix}\]

    • \(\mathrm{dim}(A)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • \(\mathrm{dim}(B)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • Is the product \(AB\) defined? -------- Yes, the matrices can be multiplied.No, the matrices cannot be multiplied.
    
    
    
    

    Solution

    Show solution
    • The dimensions of \(A\) are \(3\times2\).
    • The dimensions of \(B\) are \(2\times5\).
    • The number of columns of \(A\) is equal to the number of rows of \(B\), so we can find the product \(AB\).

    The dimensions of the product \(C=AB\) would be \(3\times5\).

    
    
65. 
    

    Question

    prevnext

    If the dimensions of matrix \(A\) are \(m\times n\) and the dimensions of matrix \(B\) are \(p\times q\), then the matrix-multiplication \(AB\) is only possible if \(n=p\).

    \[A=\begin{bmatrix} -2 & 6 & -3 & -2 & -8 & 1\\ 0 & 4 & -3 & 2 & -1 & -2\\ 6 & -9 & 3 & 3 & 6 & -4\\ -1 & 1 & 8 & -2 & -3 & -3\\ 4 & 4 & -4 & 5 & 5 & 0 \end{bmatrix}\] \[B=\begin{bmatrix} 0 & 1 & -5\\ -9 & 6 & -9\\ -4 & -3 & -6\\ -7 & -3 & -7\\ 3 & -8 & 5\\ 5 & -6 & -6 \end{bmatrix}\]

    • \(\mathrm{dim}(A)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • \(\mathrm{dim}(B)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • Is the product \(AB\) defined? -------- Yes, the matrices can be multiplied.No, the matrices cannot be multiplied.
    
    
    
    

    Solution

    Show solution
    • The dimensions of \(A\) are \(5\times6\).
    • The dimensions of \(B\) are \(6\times3\).
    • The number of columns of \(A\) is equal to the number of rows of \(B\), so we can find the product \(AB\).

    The dimensions of the product \(C=AB\) would be \(5\times3\).

    
    
66. 
    

    Question

    prevnext

    If the dimensions of matrix \(A\) are \(m\times n\) and the dimensions of matrix \(B\) are \(p\times q\), then the matrix-multiplication \(AB\) is only possible if \(n=p\).

    \[A=\begin{bmatrix} -9 & 3 & 8 & -3 & 9 & 0\\ 6 & 8 & -3 & -5 & 0 & 1\\ -9 & 4 & -7 & -7 & 2 & -1 \end{bmatrix}\] \[B=\begin{bmatrix} 0 & 7 & -6 & 0 & 0\\ -8 & 5 & 2 & -1 & -8\\ 5 & -3 & 7 & -1 & -2\\ -4 & 5 & -7 & -9 & 1\\ -8 & 6 & -3 & 9 & 7\\ -1 & 8 & 7 & 8 & 7 \end{bmatrix}\]

    • \(\mathrm{dim}(A)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • \(\mathrm{dim}(B)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • Is the product \(AB\) defined? -------- Yes, the matrices can be multiplied.No, the matrices cannot be multiplied.
    
    
    
    

    Solution

    Show solution
    • The dimensions of \(A\) are \(3\times6\).
    • The dimensions of \(B\) are \(6\times5\).
    • The number of columns of \(A\) is equal to the number of rows of \(B\), so we can find the product \(AB\).

    The dimensions of the product \(C=AB\) would be \(3\times5\).

    
    
67. 
    

    Question

    prevnext

    If the dimensions of matrix \(A\) are \(m\times n\) and the dimensions of matrix \(B\) are \(p\times q\), then the matrix-multiplication \(AB\) is only possible if \(n=p\).

    \[A=\begin{bmatrix} 3 & 9 & 3 & -9 & 6\\ 3 & -5 & -7 & 9 & 8\\ -8 & 6 & 0 & 4 & -7\\ 4 & -4 & -9 & -6 & -1 \end{bmatrix}\] \[B=\begin{bmatrix} -8 & 0\\ -1 & -1\\ -1 & 8\\ -8 & -1\\ -9 & 5 \end{bmatrix}\]

    • \(\mathrm{dim}(A)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • \(\mathrm{dim}(B)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • Is the product \(AB\) defined? -------- Yes, the matrices can be multiplied.No, the matrices cannot be multiplied.
    
    
    
    

    Solution

    Show solution
    • The dimensions of \(A\) are \(4\times5\).
    • The dimensions of \(B\) are \(5\times2\).
    • The number of columns of \(A\) is equal to the number of rows of \(B\), so we can find the product \(AB\).

    The dimensions of the product \(C=AB\) would be \(4\times2\).

    
    
68. 
    

    Question

    prevnext

    If the dimensions of matrix \(A\) are \(m\times n\) and the dimensions of matrix \(B\) are \(p\times q\), then the matrix-multiplication \(AB\) is only possible if \(n=p\).

    \[A=\begin{bmatrix} -6 & 7 & -1 & 7\\ -7 & 7 & -5 & -7\\ 3 & 7 & 3 & -8\\ 8 & 7 & 4 & 5\\ -8 & 3 & 4 & 7 \end{bmatrix}\] \[B=\begin{bmatrix} -1 & -3 & -3\\ -5 & -3 & 1\\ -3 & 4 & 8\\ 3 & 3 & 7 \end{bmatrix}\]

    • \(\mathrm{dim}(A)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • \(\mathrm{dim}(B)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • Is the product \(AB\) defined? -------- Yes, the matrices can be multiplied.No, the matrices cannot be multiplied.
    
    
    
    

    Solution

    Show solution
    • The dimensions of \(A\) are \(5\times4\).
    • The dimensions of \(B\) are \(4\times3\).
    • The number of columns of \(A\) is equal to the number of rows of \(B\), so we can find the product \(AB\).

    The dimensions of the product \(C=AB\) would be \(5\times3\).

    
    
69. 
    

    Question

    prevnext

    If the dimensions of matrix \(A\) are \(m\times n\) and the dimensions of matrix \(B\) are \(p\times q\), then the matrix-multiplication \(AB\) is only possible if \(n=p\).

    \[A=\begin{bmatrix} 5 & 7 & 8 & 4 & 0 & 0\\ 6 & 6 & -8 & -9 & 5 & -4\\ 3 & 0 & 4 & 7 & -1 & -9 \end{bmatrix}\] \[B=\begin{bmatrix} -2 & -4\\ -5 & 7\\ -8 & 6\\ -7 & -9 \end{bmatrix}\]

    • \(\mathrm{dim}(A)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • \(\mathrm{dim}(B)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • Is the product \(AB\) defined? -------- Yes, the matrices can be multiplied.No, the matrices cannot be multiplied.
    
    
    
    

    Solution

    Show solution
    • The dimensions of \(A\) are \(3\times6\).
    • The dimensions of \(B\) are \(4\times2\).
    • The number of columns of \(A\) is NOT equal to the number of rows of \(B\), so we can NOT find the product \(AB\).

    
    
70. 
    

    Question

    prevnext

    If the dimensions of matrix \(A\) are \(m\times n\) and the dimensions of matrix \(B\) are \(p\times q\), then the matrix-multiplication \(AB\) is only possible if \(n=p\).

    \[A=\begin{bmatrix} 0 & 4 & -8 & -6 & -8\\ -2 & 3 & -9 & 6 & -1\\ 1 & 1 & 2 & -9 & 1 \end{bmatrix}\] \[B=\begin{bmatrix} 9 & 0\\ 5 & -2\\ -5 & 6\\ -1 & 0\\ -4 & -1\\ 4 & -3 \end{bmatrix}\]

    • \(\mathrm{dim}(A)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • \(\mathrm{dim}(B)=\) -------- 12345678910 \(\times\)-------- 12345678910
    • Is the product \(AB\) defined? -------- Yes, the matrices can be multiplied.No, the matrices cannot be multiplied.
    
    
    
    

    Solution

    Show solution
    • The dimensions of \(A\) are \(3\times5\).
    • The dimensions of \(B\) are \(6\times2\).
    • The number of columns of \(A\) is NOT equal to the number of rows of \(B\), so we can NOT find the product \(AB\).

    
    
71. 
    

    Question

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    Matrices \(A\) and \(B\) are defined below.

    \[A = \begin{bmatrix} -1 & 0 & 1\\ 4 & 7 & 2 \end{bmatrix}\] \[B = \begin{bmatrix} 6 & -1 & 1 & 9\\ 6 & 3 & 7 & -4\\ 6 & -5 & -2 & 7 \end{bmatrix}\]

    Find the product \(C=AB\).

    \(AB=\)

    -------- -4-3-207224978	-------- -4-3-207224978	-------- -4-3-207224978	-------- -4-3-207224978
    -------- -4-3-207224978	-------- -4-3-207224978	-------- -4-3-207224978	-------- -4-3-207224978

    
    
    
    

    Solution

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    We wish to do matrix multiplication:

    \[\begin{bmatrix} -1 & 0 & 1\\ 4 & 7 & 2 \end{bmatrix}\begin{bmatrix} 6 & -1 & 1 & 9\\ 6 & 3 & 7 & -4\\ 6 & -5 & -2 & 7 \end{bmatrix}\]

    The product matrix has 2 rows and 4 columns.

    To find the product matrix’s value in row 1 and column 1, sum the products of corresponding elements of row 1 in matrix A and column 1 in matrix B. \[C_{[1,1]}=(-1)(6)+(0)(6)+(1)(6)=0\]

    To find the product matrix’s value in row 1 and column 2, sum the products of corresponding elements of row 1 in matrix A and column 2 in matrix B. \[C_{[1,2]}=(-1)(-1)+(0)(3)+(1)(-5)=-4\]

    To find the product matrix’s value in row 1 and column 3, sum the products of corresponding elements of row 1 in matrix A and column 3 in matrix B. \[C_{[1,3]}=(-1)(1)+(0)(7)+(1)(-2)=-3\]

    To find the product matrix’s value in row 1 and column 4, sum the products of corresponding elements of row 1 in matrix A and column 4 in matrix B. \[C_{[1,4]}=(-1)(9)+(0)(-4)+(1)(7)=-2\]

    To find the product matrix’s value in row 2 and column 1, sum the products of corresponding elements of row 2 in matrix A and column 1 in matrix B. \[C_{[2,1]}=(4)(6)+(7)(6)+(2)(6)=78\]

    To find the product matrix’s value in row 2 and column 2, sum the products of corresponding elements of row 2 in matrix A and column 2 in matrix B. \[C_{[2,2]}=(4)(-1)+(7)(3)+(2)(-5)=7\]

    To find the product matrix’s value in row 2 and column 3, sum the products of corresponding elements of row 2 in matrix A and column 3 in matrix B. \[C_{[2,3]}=(4)(1)+(7)(7)+(2)(-2)=49\]

    To find the product matrix’s value in row 2 and column 4, sum the products of corresponding elements of row 2 in matrix A and column 4 in matrix B. \[C_{[2,4]}=(4)(9)+(7)(-4)+(2)(7)=22\]

    All of that work leads to the answer:

    \[C = AB = \begin{bmatrix} 0 & -4 & -3 & -2\\ 78 & 7 & 49 & 22 \end{bmatrix}\]

    
    
72. 
    

    Question

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    Matrices \(A\) and \(B\) are defined below.

    \[A = \begin{bmatrix} -2 & -2 & 7\\ -3 & -7 & 4\\ -2 & 6 & 5\\ 3 & -7 & 7 \end{bmatrix}\] \[B = \begin{bmatrix} -9 & -4\\ 9 & 3\\ 0 & 8 \end{bmatrix}\]

    Find the product \(C=AB\).

    \(AB=\)

    -------- -90-36023586672	-------- -90-36023586672
    -------- -90-36023586672	-------- -90-36023586672
    -------- -90-36023586672	-------- -90-36023586672
    -------- -90-36023586672	-------- -90-36023586672

    
    
    
    

    Solution

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    We wish to do matrix multiplication:

    \[\begin{bmatrix} -2 & -2 & 7\\ -3 & -7 & 4\\ -2 & 6 & 5\\ 3 & -7 & 7 \end{bmatrix}\begin{bmatrix} -9 & -4\\ 9 & 3\\ 0 & 8 \end{bmatrix}\]

    The product matrix has 4 rows and 2 columns.

    To find the product matrix’s value in row 1 and column 1, sum the products of corresponding elements of row 1 in matrix A and column 1 in matrix B. \[C_{[1,1]}=(-2)(-9)+(-2)(9)+(7)(0)=0\]

    To find the product matrix’s value in row 1 and column 2, sum the products of corresponding elements of row 1 in matrix A and column 2 in matrix B. \[C_{[1,2]}=(-2)(-4)+(-2)(3)+(7)(8)=58\]

    To find the product matrix’s value in row 2 and column 1, sum the products of corresponding elements of row 2 in matrix A and column 1 in matrix B. \[C_{[2,1]}=(-3)(-9)+(-7)(9)+(4)(0)=-36\]

    To find the product matrix’s value in row 2 and column 2, sum the products of corresponding elements of row 2 in matrix A and column 2 in matrix B. \[C_{[2,2]}=(-3)(-4)+(-7)(3)+(4)(8)=23\]

    To find the product matrix’s value in row 3 and column 1, sum the products of corresponding elements of row 3 in matrix A and column 1 in matrix B. \[C_{[3,1]}=(-2)(-9)+(6)(9)+(5)(0)=72\]

    To find the product matrix’s value in row 3 and column 2, sum the products of corresponding elements of row 3 in matrix A and column 2 in matrix B. \[C_{[3,2]}=(-2)(-4)+(6)(3)+(5)(8)=66\]

    To find the product matrix’s value in row 4 and column 1, sum the products of corresponding elements of row 4 in matrix A and column 1 in matrix B. \[C_{[4,1]}=(3)(-9)+(-7)(9)+(7)(0)=-90\]

    To find the product matrix’s value in row 4 and column 2, sum the products of corresponding elements of row 4 in matrix A and column 2 in matrix B. \[C_{[4,2]}=(3)(-4)+(-7)(3)+(7)(8)=23\]

    All of that work leads to the answer:

    \[C = AB = \begin{bmatrix} 0 & 58\\ -36 & 23\\ 72 & 66\\ -90 & 23 \end{bmatrix}\]

    
    
73. 
    

    Question

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    Matrices \(A\) and \(B\) are defined below.

    \[A = \begin{bmatrix} 0 & 4 & -1 & 3\\ -6 & 8 & 0 & 7\\ -8 & -8 & 2 & -2 \end{bmatrix}\] \[B = \begin{bmatrix} -5 & -6\\ -8 & -4\\ 0 & -3\\ 7 & 8 \end{bmatrix}\]

    Find the product \(C=AB\).

    \(AB=\)

    -------- -111115586090	-------- -111115586090
    -------- -111115586090	-------- -111115586090
    -------- -111115586090	-------- -111115586090

    
    
    
    

    Solution

    Show solution

    We wish to do matrix multiplication:

    \[\begin{bmatrix} 0 & 4 & -1 & 3\\ -6 & 8 & 0 & 7\\ -8 & -8 & 2 & -2 \end{bmatrix}\begin{bmatrix} -5 & -6\\ -8 & -4\\ 0 & -3\\ 7 & 8 \end{bmatrix}\]

    The product matrix has 3 rows and 2 columns.

    To find the product matrix’s value in row 1 and column 1, sum the products of corresponding elements of row 1 in matrix A and column 1 in matrix B. \[C_{[1,1]}=(0)(-5)+(4)(-8)+(-1)(0)+(3)(7)=-11\]

    To find the product matrix’s value in row 1 and column 2, sum the products of corresponding elements of row 1 in matrix A and column 2 in matrix B. \[C_{[1,2]}=(0)(-6)+(4)(-4)+(-1)(-3)+(3)(8)=11\]

    To find the product matrix’s value in row 2 and column 1, sum the products of corresponding elements of row 2 in matrix A and column 1 in matrix B. \[C_{[2,1]}=(-6)(-5)+(8)(-8)+(0)(0)+(7)(7)=15\]

    To find the product matrix’s value in row 2 and column 2, sum the products of corresponding elements of row 2 in matrix A and column 2 in matrix B. \[C_{[2,2]}=(-6)(-6)+(8)(-4)+(0)(-3)+(7)(8)=60\]

    To find the product matrix’s value in row 3 and column 1, sum the products of corresponding elements of row 3 in matrix A and column 1 in matrix B. \[C_{[3,1]}=(-8)(-5)+(-8)(-8)+(2)(0)+(-2)(7)=90\]

    To find the product matrix’s value in row 3 and column 2, sum the products of corresponding elements of row 3 in matrix A and column 2 in matrix B. \[C_{[3,2]}=(-8)(-6)+(-8)(-4)+(2)(-3)+(-2)(8)=58\]

    All of that work leads to the answer:

    \[C = AB = \begin{bmatrix} -11 & 11\\ 15 & 60\\ 90 & 58 \end{bmatrix}\]

    
    
74. 
    

    Question

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    Matrices \(A\) and \(B\) are defined below.

    \[A = \begin{bmatrix} -6 & -4 & 0\\ -3 & -6 & -3\\ 1 & 0 & -2\\ 7 & -4 & 0 \end{bmatrix}\] \[B = \begin{bmatrix} -3 & -3\\ -9 & 5\\ 1 & -1 \end{bmatrix}\]

    Find the product \(C=AB\).

    \(AB=\)

    -------- -41-18-5-2-1155460	-------- -41-18-5-2-1155460
    -------- -41-18-5-2-1155460	-------- -41-18-5-2-1155460
    -------- -41-18-5-2-1155460	-------- -41-18-5-2-1155460
    -------- -41-18-5-2-1155460	-------- -41-18-5-2-1155460

    
    
    
    

    Solution

    Show solution

    We wish to do matrix multiplication:

    \[\begin{bmatrix} -6 & -4 & 0\\ -3 & -6 & -3\\ 1 & 0 & -2\\ 7 & -4 & 0 \end{bmatrix}\begin{bmatrix} -3 & -3\\ -9 & 5\\ 1 & -1 \end{bmatrix}\]

    The product matrix has 4 rows and 2 columns.

    To find the product matrix’s value in row 1 and column 1, sum the products of corresponding elements of row 1 in matrix A and column 1 in matrix B. \[C_{[1,1]}=(-6)(-3)+(-4)(-9)+(0)(1)=54\]

    To find the product matrix’s value in row 1 and column 2, sum the products of corresponding elements of row 1 in matrix A and column 2 in matrix B. \[C_{[1,2]}=(-6)(-3)+(-4)(5)+(0)(-1)=-2\]

    To find the product matrix’s value in row 2 and column 1, sum the products of corresponding elements of row 2 in matrix A and column 1 in matrix B. \[C_{[2,1]}=(-3)(-3)+(-6)(-9)+(-3)(1)=60\]

    To find the product matrix’s value in row 2 and column 2, sum the products of corresponding elements of row 2 in matrix A and column 2 in matrix B. \[C_{[2,2]}=(-3)(-3)+(-6)(5)+(-3)(-1)=-18\]

    To find the product matrix’s value in row 3 and column 1, sum the products of corresponding elements of row 3 in matrix A and column 1 in matrix B. \[C_{[3,1]}=(1)(-3)+(0)(-9)+(-2)(1)=-5\]

    To find the product matrix’s value in row 3 and column 2, sum the products of corresponding elements of row 3 in matrix A and column 2 in matrix B. \[C_{[3,2]}=(1)(-3)+(0)(5)+(-2)(-1)=-1\]

    To find the product matrix’s value in row 4 and column 1, sum the products of corresponding elements of row 4 in matrix A and column 1 in matrix B. \[C_{[4,1]}=(7)(-3)+(-4)(-9)+(0)(1)=15\]

    To find the product matrix’s value in row 4 and column 2, sum the products of corresponding elements of row 4 in matrix A and column 2 in matrix B. \[C_{[4,2]}=(7)(-3)+(-4)(5)+(0)(-1)=-41\]

    All of that work leads to the answer:

    \[C = AB = \begin{bmatrix} 54 & -2\\ 60 & -18\\ -5 & -1\\ 15 & -41 \end{bmatrix}\]

    
    
75. 
    

    Question

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    Matrices \(A\) and \(B\) are defined below.

    \[A = \begin{bmatrix} -2 & 4 & 8 & -3\\ -4 & -3 & 3 & 5\\ 7 & -1 & 1 & 9 \end{bmatrix}\] \[B = \begin{bmatrix} -3 & -4\\ 5 & -8\\ -7 & 1\\ 8 & -6 \end{bmatrix}\]

    Find the product \(C=AB\).

    \(AB=\)

    -------- -73-542131639	-------- -73-542131639
    -------- -73-542131639	-------- -73-542131639
    -------- -73-542131639	-------- -73-542131639

    
    
    
    

    Solution

    Show solution

    We wish to do matrix multiplication:

    \[\begin{bmatrix} -2 & 4 & 8 & -3\\ -4 & -3 & 3 & 5\\ 7 & -1 & 1 & 9 \end{bmatrix}\begin{bmatrix} -3 & -4\\ 5 & -8\\ -7 & 1\\ 8 & -6 \end{bmatrix}\]

    The product matrix has 3 rows and 2 columns.

    To find the product matrix’s value in row 1 and column 1, sum the products of corresponding elements of row 1 in matrix A and column 1 in matrix B. \[C_{[1,1]}=(-2)(-3)+(4)(5)+(8)(-7)+(-3)(8)=-54\]

    To find the product matrix’s value in row 1 and column 2, sum the products of corresponding elements of row 1 in matrix A and column 2 in matrix B. \[C_{[1,2]}=(-2)(-4)+(4)(-8)+(8)(1)+(-3)(-6)=2\]

    To find the product matrix’s value in row 2 and column 1, sum the products of corresponding elements of row 2 in matrix A and column 1 in matrix B. \[C_{[2,1]}=(-4)(-3)+(-3)(5)+(3)(-7)+(5)(8)=16\]

    To find the product matrix’s value in row 2 and column 2, sum the products of corresponding elements of row 2 in matrix A and column 2 in matrix B. \[C_{[2,2]}=(-4)(-4)+(-3)(-8)+(3)(1)+(5)(-6)=13\]

    To find the product matrix’s value in row 3 and column 1, sum the products of corresponding elements of row 3 in matrix A and column 1 in matrix B. \[C_{[3,1]}=(7)(-3)+(-1)(5)+(1)(-7)+(9)(8)=39\]

    To find the product matrix’s value in row 3 and column 2, sum the products of corresponding elements of row 3 in matrix A and column 2 in matrix B. \[C_{[3,2]}=(7)(-4)+(-1)(-8)+(1)(1)+(9)(-6)=-73\]

    All of that work leads to the answer:

    \[C = AB = \begin{bmatrix} -54 & 2\\ 16 & 13\\ 39 & -73 \end{bmatrix}\]

    
    
76. 
    

    Question

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    Matrices \(A\) and \(B\) are defined below.

    \[A = \begin{bmatrix} 0 & 9 & 7\\ 2 & 8 & 9\\ -1 & -4 & 9\\ -3 & -9 & 8 \end{bmatrix}\] \[B = \begin{bmatrix} -9 & 5\\ 7 & -3\\ -1 & -7 \end{bmatrix}\]

    Find the product \(C=AB\).

    \(AB=\)

    -------- -77-76-56-44-282956	-------- -77-76-56-44-282956
    -------- -77-76-56-44-282956	-------- -77-76-56-44-282956
    -------- -77-76-56-44-282956	-------- -77-76-56-44-282956
    -------- -77-76-56-44-282956	-------- -77-76-56-44-282956

    
    
    
    

    Solution

    Show solution

    We wish to do matrix multiplication:

    \[\begin{bmatrix} 0 & 9 & 7\\ 2 & 8 & 9\\ -1 & -4 & 9\\ -3 & -9 & 8 \end{bmatrix}\begin{bmatrix} -9 & 5\\ 7 & -3\\ -1 & -7 \end{bmatrix}\]

    The product matrix has 4 rows and 2 columns.

    To find the product matrix’s value in row 1 and column 1, sum the products of corresponding elements of row 1 in matrix A and column 1 in matrix B. \[C_{[1,1]}=(0)(-9)+(9)(7)+(7)(-1)=56\]

    To find the product matrix’s value in row 1 and column 2, sum the products of corresponding elements of row 1 in matrix A and column 2 in matrix B. \[C_{[1,2]}=(0)(5)+(9)(-3)+(7)(-7)=-76\]

    To find the product matrix’s value in row 2 and column 1, sum the products of corresponding elements of row 2 in matrix A and column 1 in matrix B. \[C_{[2,1]}=(2)(-9)+(8)(7)+(9)(-1)=29\]

    To find the product matrix’s value in row 2 and column 2, sum the products of corresponding elements of row 2 in matrix A and column 2 in matrix B. \[C_{[2,2]}=(2)(5)+(8)(-3)+(9)(-7)=-77\]

    To find the product matrix’s value in row 3 and column 1, sum the products of corresponding elements of row 3 in matrix A and column 1 in matrix B. \[C_{[3,1]}=(-1)(-9)+(-4)(7)+(9)(-1)=-28\]

    To find the product matrix’s value in row 3 and column 2, sum the products of corresponding elements of row 3 in matrix A and column 2 in matrix B. \[C_{[3,2]}=(-1)(5)+(-4)(-3)+(9)(-7)=-56\]

    To find the product matrix’s value in row 4 and column 1, sum the products of corresponding elements of row 4 in matrix A and column 1 in matrix B. \[C_{[4,1]}=(-3)(-9)+(-9)(7)+(8)(-1)=-44\]

    To find the product matrix’s value in row 4 and column 2, sum the products of corresponding elements of row 4 in matrix A and column 2 in matrix B. \[C_{[4,2]}=(-3)(5)+(-9)(-3)+(8)(-7)=-44\]

    All of that work leads to the answer:

    \[C = AB = \begin{bmatrix} 56 & -76\\ 29 & -77\\ -28 & -56\\ -44 & -44 \end{bmatrix}\]

    
    
77. 
    

    Question

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    Matrices \(A\) and \(B\) are defined below.

    \[A = \begin{bmatrix} 5 & -3 & -2\\ -3 & -9 & 7\\ 8 & 7 & -9\\ 7 & -8 & 2 \end{bmatrix}\] \[B = \begin{bmatrix} -1 & 4\\ -1 & 3\\ 1 & 8 \end{bmatrix}\]

    Find the product \(C=AB\).

    \(AB=\)

    -------- -24-19-5-43171920	-------- -24-19-5-43171920
    -------- -24-19-5-43171920	-------- -24-19-5-43171920
    -------- -24-19-5-43171920	-------- -24-19-5-43171920
    -------- -24-19-5-43171920	-------- -24-19-5-43171920

    
    
    
    

    Solution

    Show solution

    We wish to do matrix multiplication:

    \[\begin{bmatrix} 5 & -3 & -2\\ -3 & -9 & 7\\ 8 & 7 & -9\\ 7 & -8 & 2 \end{bmatrix}\begin{bmatrix} -1 & 4\\ -1 & 3\\ 1 & 8 \end{bmatrix}\]

    The product matrix has 4 rows and 2 columns.

    To find the product matrix’s value in row 1 and column 1, sum the products of corresponding elements of row 1 in matrix A and column 1 in matrix B. \[C_{[1,1]}=(5)(-1)+(-3)(-1)+(-2)(1)=-4\]

    To find the product matrix’s value in row 1 and column 2, sum the products of corresponding elements of row 1 in matrix A and column 2 in matrix B. \[C_{[1,2]}=(5)(4)+(-3)(3)+(-2)(8)=-5\]

    To find the product matrix’s value in row 2 and column 1, sum the products of corresponding elements of row 2 in matrix A and column 1 in matrix B. \[C_{[2,1]}=(-3)(-1)+(-9)(-1)+(7)(1)=19\]

    To find the product matrix’s value in row 2 and column 2, sum the products of corresponding elements of row 2 in matrix A and column 2 in matrix B. \[C_{[2,2]}=(-3)(4)+(-9)(3)+(7)(8)=17\]

    To find the product matrix’s value in row 3 and column 1, sum the products of corresponding elements of row 3 in matrix A and column 1 in matrix B. \[C_{[3,1]}=(8)(-1)+(7)(-1)+(-9)(1)=-24\]

    To find the product matrix’s value in row 3 and column 2, sum the products of corresponding elements of row 3 in matrix A and column 2 in matrix B. \[C_{[3,2]}=(8)(4)+(7)(3)+(-9)(8)=-19\]

    To find the product matrix’s value in row 4 and column 1, sum the products of corresponding elements of row 4 in matrix A and column 1 in matrix B. \[C_{[4,1]}=(7)(-1)+(-8)(-1)+(2)(1)=3\]

    To find the product matrix’s value in row 4 and column 2, sum the products of corresponding elements of row 4 in matrix A and column 2 in matrix B. \[C_{[4,2]}=(7)(4)+(-8)(3)+(2)(8)=20\]

    All of that work leads to the answer:

    \[C = AB = \begin{bmatrix} -4 & -5\\ 19 & 17\\ -24 & -19\\ 3 & 20 \end{bmatrix}\]

    
    
78. 
    

    Question

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    Matrices \(A\) and \(B\) are defined below.

    \[A = \begin{bmatrix} -2 & 6\\ 5 & 0\\ -1 & 1\\ -1 & 3 \end{bmatrix}\] \[B = \begin{bmatrix} 0 & -5 & 9\\ -1 & -7 & 6 \end{bmatrix}\]

    Find the product \(C=AB\).

    \(AB=\)

    -------- -32-25-16-6-3-2-1091845	-------- -32-25-16-6-3-2-1091845	-------- -32-25-16-6-3-2-1091845
    -------- -32-25-16-6-3-2-1091845	-------- -32-25-16-6-3-2-1091845	-------- -32-25-16-6-3-2-1091845
    -------- -32-25-16-6-3-2-1091845	-------- -32-25-16-6-3-2-1091845	-------- -32-25-16-6-3-2-1091845
    -------- -32-25-16-6-3-2-1091845	-------- -32-25-16-6-3-2-1091845	-------- -32-25-16-6-3-2-1091845

    
    
    
    

    Solution

    Show solution

    We wish to do matrix multiplication:

    \[\begin{bmatrix} -2 & 6\\ 5 & 0\\ -1 & 1\\ -1 & 3 \end{bmatrix}\begin{bmatrix} 0 & -5 & 9\\ -1 & -7 & 6 \end{bmatrix}\]

    The product matrix has 4 rows and 3 columns.

    To find the product matrix’s value in row 1 and column 1, sum the products of corresponding elements of row 1 in matrix A and column 1 in matrix B. \[C_{[1,1]}=(-2)(0)+(6)(-1)=-6\]

    To find the product matrix’s value in row 1 and column 2, sum the products of corresponding elements of row 1 in matrix A and column 2 in matrix B. \[C_{[1,2]}=(-2)(-5)+(6)(-7)=-32\]

    To find the product matrix’s value in row 1 and column 3, sum the products of corresponding elements of row 1 in matrix A and column 3 in matrix B. \[C_{[1,3]}=(-2)(9)+(6)(6)=18\]

    To find the product matrix’s value in row 2 and column 1, sum the products of corresponding elements of row 2 in matrix A and column 1 in matrix B. \[C_{[2,1]}=(5)(0)+(0)(-1)=0\]

    To find the product matrix’s value in row 2 and column 2, sum the products of corresponding elements of row 2 in matrix A and column 2 in matrix B. \[C_{[2,2]}=(5)(-5)+(0)(-7)=-25\]

    To find the product matrix’s value in row 2 and column 3, sum the products of corresponding elements of row 2 in matrix A and column 3 in matrix B. \[C_{[2,3]}=(5)(9)+(0)(6)=45\]

    To find the product matrix’s value in row 3 and column 1, sum the products of corresponding elements of row 3 in matrix A and column 1 in matrix B. \[C_{[3,1]}=(-1)(0)+(1)(-1)=-1\]

    To find the product matrix’s value in row 3 and column 2, sum the products of corresponding elements of row 3 in matrix A and column 2 in matrix B. \[C_{[3,2]}=(-1)(-5)+(1)(-7)=-2\]

    To find the product matrix’s value in row 3 and column 3, sum the products of corresponding elements of row 3 in matrix A and column 3 in matrix B. \[C_{[3,3]}=(-1)(9)+(1)(6)=-3\]

    To find the product matrix’s value in row 4 and column 1, sum the products of corresponding elements of row 4 in matrix A and column 1 in matrix B. \[C_{[4,1]}=(-1)(0)+(3)(-1)=-3\]

    To find the product matrix’s value in row 4 and column 2, sum the products of corresponding elements of row 4 in matrix A and column 2 in matrix B. \[C_{[4,2]}=(-1)(-5)+(3)(-7)=-16\]

    To find the product matrix’s value in row 4 and column 3, sum the products of corresponding elements of row 4 in matrix A and column 3 in matrix B. \[C_{[4,3]}=(-1)(9)+(3)(6)=9\]

    All of that work leads to the answer:

    \[C = AB = \begin{bmatrix} -6 & -32 & 18\\ 0 & -25 & 45\\ -1 & -2 & -3\\ -3 & -16 & 9 \end{bmatrix}\]

    
    
79. 
    

    Question

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    Matrices \(A\) and \(B\) are defined below.

    \[A = \begin{bmatrix} -9 & 0\\ 0 & 2\\ 0 & -1 \end{bmatrix}\] \[B = \begin{bmatrix} 6 & -6 & 3 & 7\\ -8 & -5 & 0 & 4 \end{bmatrix}\]

    Find the product \(C=AB\).

    \(AB=\)

    -------- -63-54-27-16-10-405854	-------- -63-54-27-16-10-405854	-------- -63-54-27-16-10-405854	-------- -63-54-27-16-10-405854
    -------- -63-54-27-16-10-405854	-------- -63-54-27-16-10-405854	-------- -63-54-27-16-10-405854	-------- -63-54-27-16-10-405854
    -------- -63-54-27-16-10-405854	-------- -63-54-27-16-10-405854	-------- -63-54-27-16-10-405854	-------- -63-54-27-16-10-405854

    
    
    
    

    Solution

    Show solution

    We wish to do matrix multiplication:

    \[\begin{bmatrix} -9 & 0\\ 0 & 2\\ 0 & -1 \end{bmatrix}\begin{bmatrix} 6 & -6 & 3 & 7\\ -8 & -5 & 0 & 4 \end{bmatrix}\]

    The product matrix has 3 rows and 4 columns.

    To find the product matrix’s value in row 1 and column 1, sum the products of corresponding elements of row 1 in matrix A and column 1 in matrix B. \[C_{[1,1]}=(-9)(6)+(0)(-8)=-54\]

    To find the product matrix’s value in row 1 and column 2, sum the products of corresponding elements of row 1 in matrix A and column 2 in matrix B. \[C_{[1,2]}=(-9)(-6)+(0)(-5)=54\]

    To find the product matrix’s value in row 1 and column 3, sum the products of corresponding elements of row 1 in matrix A and column 3 in matrix B. \[C_{[1,3]}=(-9)(3)+(0)(0)=-27\]

    To find the product matrix’s value in row 1 and column 4, sum the products of corresponding elements of row 1 in matrix A and column 4 in matrix B. \[C_{[1,4]}=(-9)(7)+(0)(4)=-63\]

    To find the product matrix’s value in row 2 and column 1, sum the products of corresponding elements of row 2 in matrix A and column 1 in matrix B. \[C_{[2,1]}=(0)(6)+(2)(-8)=-16\]

    To find the product matrix’s value in row 2 and column 2, sum the products of corresponding elements of row 2 in matrix A and column 2 in matrix B. \[C_{[2,2]}=(0)(-6)+(2)(-5)=-10\]

    To find the product matrix’s value in row 2 and column 3, sum the products of corresponding elements of row 2 in matrix A and column 3 in matrix B. \[C_{[2,3]}=(0)(3)+(2)(0)=0\]

    To find the product matrix’s value in row 2 and column 4, sum the products of corresponding elements of row 2 in matrix A and column 4 in matrix B. \[C_{[2,4]}=(0)(7)+(2)(4)=8\]

    To find the product matrix’s value in row 3 and column 1, sum the products of corresponding elements of row 3 in matrix A and column 1 in matrix B. \[C_{[3,1]}=(0)(6)+(-1)(-8)=8\]

    To find the product matrix’s value in row 3 and column 2, sum the products of corresponding elements of row 3 in matrix A and column 2 in matrix B. \[C_{[3,2]}=(0)(-6)+(-1)(-5)=5\]

    To find the product matrix’s value in row 3 and column 3, sum the products of corresponding elements of row 3 in matrix A and column 3 in matrix B. \[C_{[3,3]}=(0)(3)+(-1)(0)=0\]

    To find the product matrix’s value in row 3 and column 4, sum the products of corresponding elements of row 3 in matrix A and column 4 in matrix B. \[C_{[3,4]}=(0)(7)+(-1)(4)=-4\]

    All of that work leads to the answer:

    \[C = AB = \begin{bmatrix} -54 & 54 & -27 & -63\\ -16 & -10 & 0 & 8\\ 8 & 5 & 0 & -4 \end{bmatrix}\]

    
    
80. 
    

    Question

    prevnext

    Matrices \(A\) and \(B\) are defined below.

    \[A = \begin{bmatrix} 9 & -3 & 8\\ -7 & 3 & -6\\ 4 & -3 & -5\\ 7 & 0 & -9 \end{bmatrix}\] \[B = \begin{bmatrix} 1 & -6\\ 6 & -7\\ 4 & -7 \end{bmatrix}\]

    Find the product \(C=AB\).

    \(AB=\)

    -------- -89-34-29-1321233263	-------- -89-34-29-1321233263
    -------- -89-34-29-1321233263	-------- -89-34-29-1321233263
    -------- -89-34-29-1321233263	-------- -89-34-29-1321233263
    -------- -89-34-29-1321233263	-------- -89-34-29-1321233263

    
    
    
    

    Solution

    Show solution

    We wish to do matrix multiplication:

    \[\begin{bmatrix} 9 & -3 & 8\\ -7 & 3 & -6\\ 4 & -3 & -5\\ 7 & 0 & -9 \end{bmatrix}\begin{bmatrix} 1 & -6\\ 6 & -7\\ 4 & -7 \end{bmatrix}\]

    The product matrix has 4 rows and 2 columns.

    To find the product matrix’s value in row 1 and column 1, sum the products of corresponding elements of row 1 in matrix A and column 1 in matrix B. \[C_{[1,1]}=(9)(1)+(-3)(6)+(8)(4)=23\]

    To find the product matrix’s value in row 1 and column 2, sum the products of corresponding elements of row 1 in matrix A and column 2 in matrix B. \[C_{[1,2]}=(9)(-6)+(-3)(-7)+(8)(-7)=-89\]

    To find the product matrix’s value in row 2 and column 1, sum the products of corresponding elements of row 2 in matrix A and column 1 in matrix B. \[C_{[2,1]}=(-7)(1)+(3)(6)+(-6)(4)=-13\]

    To find the product matrix’s value in row 2 and column 2, sum the products of corresponding elements of row 2 in matrix A and column 2 in matrix B. \[C_{[2,2]}=(-7)(-6)+(3)(-7)+(-6)(-7)=63\]

    To find the product matrix’s value in row 3 and column 1, sum the products of corresponding elements of row 3 in matrix A and column 1 in matrix B. \[C_{[3,1]}=(4)(1)+(-3)(6)+(-5)(4)=-34\]

    To find the product matrix’s value in row 3 and column 2, sum the products of corresponding elements of row 3 in matrix A and column 2 in matrix B. \[C_{[3,2]}=(4)(-6)+(-3)(-7)+(-5)(-7)=32\]

    To find the product matrix’s value in row 4 and column 1, sum the products of corresponding elements of row 4 in matrix A and column 1 in matrix B. \[C_{[4,1]}=(7)(1)+(0)(6)+(-9)(4)=-29\]

    To find the product matrix’s value in row 4 and column 2, sum the products of corresponding elements of row 4 in matrix A and column 2 in matrix B. \[C_{[4,2]}=(7)(-6)+(0)(-7)+(-9)(-7)=21\]

    All of that work leads to the answer:

    \[C = AB = \begin{bmatrix} 23 & -89\\ -13 & 63\\ -34 & 32\\ -29 & 21 \end{bmatrix}\]

    
    
81. 
    

    Question

    prevnext

    In a 15-minute city, all the following grocery stores are within a quick walk, and their prices (in USD) for some foods are shown.

    	apple	banana	cherry	durian	eggplant	fig	grapefruit	honeydew
    Lovely Lunches	1.19	0.84	6.24	9.22	4.01	3.70	2.28	4.79
    More Munchies	1.70	1.06	4.92	9.15	3.01	3.10	3.19	5.75
    Nick’s Necessities	1.52	0.81	5.12	7.53	3.13	4.95	2.33	6.97
    Overpriced Options	1.67	1.01	6.47	9.50	4.39	3.73	2.78	6.32
    Price Pincher	1.65	0.97	6.60	6.90	3.41	4.34	2.88	4.66

    Each of the people below plans to go shopping for those foods. Their grocery lists are shown.

    	Quintin	Rachel	Sebastian	Trevor	Ursula	Venus
    apple	0	3	2	1	1	0
    banana	1	4	0	0	2	0
    cherry	0	1	4	0	2	1
    durian	1	2	0	4	0	4
    eggplant	0	0	2	2	4	2
    fig	3	1	0	3	0	0
    grapefruit	2	2	1	2	2	2
    honeydew	1	0	2	0	0	0

    We can create these matrices in SageMath using the following commands:

    P = matrix([[1.19,0.84,6.24,9.22,4.01,3.70,2.28,4.79],
                [1.70,1.06,4.92,9.15,3.01,3.10,3.19,5.75],
                [1.52,0.81,5.12,7.53,3.13,4.95,2.33,6.97],
                [1.67,1.01,6.47,9.50,4.39,3.73,2.78,6.32],
                [1.65,0.97,6.60,6.90,3.41,4.34,2.88,4.66]])
    
    Q = matrix([[0,3,2,1,1,0],
                [1,4,0,0,2,0],
                [0,1,4,0,2,1],
                [1,2,0,4,0,4],
                [0,0,2,2,4,2],
                [3,1,0,3,0,0],
                [2,2,1,2,2,2],
                [1,0,2,0,0,0]])

    We can find the product of these matrices.

    pretty_print(P*Q)

    What is the maximum possible cost of a grocery trip?

    
    

    Solution

    Show solution

    The maximum cost is 65.2 USD. When you run the lines of code, you should get the following matrix:

    \[\begin{bmatrix} 30.51 & 39.87 & 47.22 & 61.75 & 35.95 & 55.7\\ 31.64 & 42.04 & 43.79 & 60 & 32.08 & 53.92\\ 34.82 & 37.59 & 46.05 & 57.41 & 30.56 & 46.16\\ 33.58 & 43.81 & 53.42 & 65.2 & 39.75 & 58.81\\ 31.31 & 39.33 & 48.72 & 54.85 & 36.19 & 46.78 \end{bmatrix}\] If there are many trailing 0s, you can use the n function to get numerical approximations.

    P = matrix([[1.19,0.84,6.24,9.22,4.01,3.70,2.28,4.79],
                [1.70,1.06,4.92,9.15,3.01,3.10,3.19,5.75],
                [1.52,0.81,5.12,7.53,3.13,4.95,2.33,6.97],
                [1.67,1.01,6.47,9.50,4.39,3.73,2.78,6.32],
                [1.65,0.97,6.60,6.90,3.41,4.34,2.88,4.66]])
    Q = matrix([[0,3,2,1,1,0],
                [1,4,0,0,2,0],
                [0,1,4,0,2,1],
                [1,2,0,4,0,4],
                [0,0,2,2,4,2],
                [3,1,0,3,0,0],
                [2,2,1,2,2,2],
                [1,0,2,0,0,0]])
    C = n(P*Q,digits=4)
    pretty_print(C)

    You can manually identify the maximum cost.

    Getting sagemath to identify the largest element is annoyingly unintuitive.

    print(max(C.list()))

    
    
82. 
    

    Question

    prevnext

    In a 15-minute city, all the following grocery stores are within a quick walk, and their prices (in USD) for some foods are shown.

    	apple	banana	cherry	durian	eggplant	fig	grapefruit	honeydew
    Lovely Lunches	1.32	1.12	5.14	6.32	4.29	4.23	3.11	4.80
    More Munchies	1.42	0.91	7.23	8.07	3.74	4.63	2.52	4.50
    Nick’s Necessities	1.65	0.91	4.76	7.80	3.38	3.06	3.32	6.98
    Overpriced Options	1.78	1.01	6.84	6.16	4.97	4.06	2.83	6.52
    Price Pincher	1.61	0.70	6.17	8.39	4.46	4.29	3.04	5.96

    Each of the people below plans to go shopping for those foods. Their grocery lists are shown.

    	Quintin	Rachel	Sebastian	Trevor	Ursula	Venus
    apple	1	1	0	1	1	4
    banana	0	1	2	0	4	0
    cherry	0	0	0	3	2	1
    durian	0	0	0	0	1	2
    eggplant	0	0	1	2	0	0
    fig	4	2	4	4	0	1
    grapefruit	3	1	2	2	2	0
    honeydew	1	4	0	2	0	1

    We can create these matrices in SageMath using the following commands:

    P = matrix([[1.32,1.12,5.14,6.32,4.29,4.23,3.11,4.80],
                [1.42,0.91,7.23,8.07,3.74,4.63,2.52,4.50],
                [1.65,0.91,4.76,7.80,3.38,3.06,3.32,6.98],
                [1.78,1.01,6.84,6.16,4.97,4.06,2.83,6.52],
                [1.61,0.70,6.17,8.39,4.46,4.29,3.04,5.96]])
    
    Q = matrix([[1,1,0,1,1,4],
                [0,1,2,0,4,0],
                [0,0,0,3,2,1],
                [0,0,0,0,1,2],
                [0,0,1,2,0,0],
                [4,2,4,4,0,1],
                [3,1,2,2,2,0],
                [1,4,0,2,0,1]])

    We can find the product of these matrices.

    pretty_print(P*Q)

    What is the maximum possible cost of a grocery trip?

    
    

    Solution

    Show solution

    The maximum cost is 67.18 USD. When you run the lines of code, you should get the following matrix:

    \[\begin{bmatrix} 32.37 & 33.21 & 29.67 & 58.06 & 28.62 & 32.09\\ 32 & 32.11 & 29.12 & 63.15 & 32.63 & 38.18\\ 30.83 & 39.92 & 24.08 & 55.53 & 29.25 & 37\\ 33.03 & 39.82 & 28.89 & 67.18 & 31.32 & 36.86\\ 33.85 & 37.77 & 29.1 & 64.2 & 31.22 & 39.64 \end{bmatrix}\] If there are many trailing 0s, you can use the n function to get numerical approximations.

    P = matrix([[1.32,1.12,5.14,6.32,4.29,4.23,3.11,4.80],
                [1.42,0.91,7.23,8.07,3.74,4.63,2.52,4.50],
                [1.65,0.91,4.76,7.80,3.38,3.06,3.32,6.98],
                [1.78,1.01,6.84,6.16,4.97,4.06,2.83,6.52],
                [1.61,0.70,6.17,8.39,4.46,4.29,3.04,5.96]])
    Q = matrix([[1,1,0,1,1,4],
                [0,1,2,0,4,0],
                [0,0,0,3,2,1],
                [0,0,0,0,1,2],
                [0,0,1,2,0,0],
                [4,2,4,4,0,1],
                [3,1,2,2,2,0],
                [1,4,0,2,0,1]])
    C = n(P*Q,digits=4)
    pretty_print(C)

    You can manually identify the maximum cost.

    Getting sagemath to identify the largest element is annoyingly unintuitive.

    print(max(C.list()))

    
    
83. 
    

    Question

    prevnext

    In a 15-minute city, all the following grocery stores are within a quick walk, and their prices (in USD) for some foods are shown.

    	apple	banana	cherry	durian	eggplant	fig	grapefruit	honeydew
    Lovely Lunches	1.79	0.85	7.24	9.56	4.73	4.92	2.96	4.88
    More Munchies	1.30	1.02	5.59	7.41	4.09	4.66	3.66	4.75
    Nick’s Necessities	1.23	0.77	7.04	9.19	3.63	4.30	2.27	5.79
    Overpriced Options	1.76	0.97	5.67	9.56	3.29	3.84	2.73	5.31
    Price Pincher	1.55	0.76	6.73	8.27	3.76	3.42	2.74	5.16

    Each of the people below plans to go shopping for those foods. Their grocery lists are shown.

    	Quintin	Rachel	Sebastian	Trevor	Ursula	Venus
    apple	4	4	2	4	1	2
    banana	0	0	2	1	3	2
    cherry	0	0	1	0	0	1
    durian	2	2	0	0	2	0
    eggplant	2	2	2	0	0	4
    fig	3	3	1	2	1	1
    grapefruit	1	1	0	1	4	1
    honeydew	1	0	3	2	1	0

    We can create these matrices in SageMath using the following commands:

    P = matrix([[1.79,0.85,7.24,9.56,4.73,4.92,2.96,4.88],
                [1.30,1.02,5.59,7.41,4.09,4.66,3.66,4.75],
                [1.23,0.77,7.04,9.19,3.63,4.30,2.27,5.79],
                [1.76,0.97,5.67,9.56,3.29,3.84,2.73,5.31],
                [1.55,0.76,6.73,8.27,3.76,3.42,2.74,5.16]])
    
    Q = matrix([[4,4,2,4,1,2],
                [0,0,2,1,3,2],
                [0,0,1,0,0,1],
                [2,2,0,0,2,0],
                [2,2,2,0,0,4],
                [3,3,1,2,1,1],
                [1,1,0,1,4,1],
                [1,0,3,2,1,0]])

    We can find the product of these matrices.

    pretty_print(P*Q)

    What is the maximum possible cost of a grocery trip?

    
    

    Solution

    Show solution

    The maximum cost is 58.34 USD. When you run the lines of code, you should get the following matrix:

    \[\begin{bmatrix} 58.34 & 53.46 & 41.54 & 30.57 & 45.1 & 39.32\\ 50.59 & 45.84 & 37.32 & 28.7 & 43.23 & 34.91\\ 51.52 & 45.73 & 39.97 & 28.14 & 41.09 & 32.13\\ 52.3 & 46.99 & 37.48 & 29.04 & 43.86 & 30.86\\ 48.42 & 43.26 & 37.77 & 26.86 & 39.91 & 32.55 \end{bmatrix}\] If there are many trailing 0s, you can use the n function to get numerical approximations.

    P = matrix([[1.79,0.85,7.24,9.56,4.73,4.92,2.96,4.88],
                [1.30,1.02,5.59,7.41,4.09,4.66,3.66,4.75],
                [1.23,0.77,7.04,9.19,3.63,4.30,2.27,5.79],
                [1.76,0.97,5.67,9.56,3.29,3.84,2.73,5.31],
                [1.55,0.76,6.73,8.27,3.76,3.42,2.74,5.16]])
    Q = matrix([[4,4,2,4,1,2],
                [0,0,2,1,3,2],
                [0,0,1,0,0,1],
                [2,2,0,0,2,0],
                [2,2,2,0,0,4],
                [3,3,1,2,1,1],
                [1,1,0,1,4,1],
                [1,0,3,2,1,0]])
    C = n(P*Q,digits=4)
    pretty_print(C)

    You can manually identify the maximum cost.

    Getting sagemath to identify the largest element is annoyingly unintuitive.

    print(max(C.list()))

    
    
84. 
    

    Question

    prevnext

    In a 15-minute city, all the following grocery stores are within a quick walk, and their prices (in USD) for some foods are shown.

    	apple	banana	cherry	durian	eggplant	fig	grapefruit	honeydew
    Lovely Lunches	1.36	0.84	4.60	7.59	3.16	3.60	2.27	5.65
    More Munchies	1.38	0.99	4.86	8.83	4.54	4.50	2.54	6.01
    Nick’s Necessities	1.14	0.92	7.38	7.74	4.90	4.26	2.29	5.69
    Overpriced Options	1.22	0.85	6.79	7.52	3.17	3.24	2.49	5.76
    Price Pincher	1.85	0.70	4.55	9.10	4.35	3.79	2.39	4.64

    Each of the people below plans to go shopping for those foods. Their grocery lists are shown.

    	Quintin	Rachel	Sebastian	Trevor	Ursula	Venus
    apple	0	2	2	1	4	0
    banana	4	0	2	4	1	1
    cherry	1	0	1	2	1	2
    durian	1	2	0	1	0	2
    eggplant	0	1	0	2	2	3
    fig	0	1	1	1	3	1
    grapefruit	0	0	4	0	0	1
    honeydew	2	1	1	2	1	4

    We can create these matrices in SageMath using the following commands:

    P = matrix([[1.36,0.84,4.60,7.59,3.16,3.60,2.27,5.65],
                [1.38,0.99,4.86,8.83,4.54,4.50,2.54,6.01],
                [1.14,0.92,7.38,7.74,4.90,4.26,2.29,5.69],
                [1.22,0.85,6.79,7.52,3.17,3.24,2.49,5.76],
                [1.85,0.70,4.55,9.10,4.35,3.79,2.39,4.64]])
    
    Q = matrix([[0,2,2,1,4,0],
                [4,0,2,4,1,1],
                [1,0,1,2,1,2],
                [1,2,0,1,0,2],
                [0,1,0,2,2,3],
                [0,1,1,1,3,1],
                [0,0,4,0,0,1],
                [2,1,1,2,1,4]])

    We can find the product of these matrices.

    pretty_print(P*Q)

    What is the maximum possible cost of a grocery trip?

    
    

    Solution

    Show solution

    The maximum cost is 75.17 USD. When you run the lines of code, you should get the following matrix:

    \[\begin{bmatrix} 26.85 & 30.31 & 27.33 & 42.73 & 33.65 & 63.17\\ 29.67 & 35.47 & 30.27 & 49.49 & 39.96 & 73.07\\ 30.18 & 32.61 & 30.61 & 52.76 & 41.13 & 75.17\\ 29.23 & 29.65 & 29.89 & 46.82 & 34.34 & 67.75\\ 25.73 & 34.68 & 27.64 & 44.62 & 37.36 & 65.79 \end{bmatrix}\] If there are many trailing 0s, you can use the n function to get numerical approximations.

    P = matrix([[1.36,0.84,4.60,7.59,3.16,3.60,2.27,5.65],
                [1.38,0.99,4.86,8.83,4.54,4.50,2.54,6.01],
                [1.14,0.92,7.38,7.74,4.90,4.26,2.29,5.69],
                [1.22,0.85,6.79,7.52,3.17,3.24,2.49,5.76],
                [1.85,0.70,4.55,9.10,4.35,3.79,2.39,4.64]])
    Q = matrix([[0,2,2,1,4,0],
                [4,0,2,4,1,1],
                [1,0,1,2,1,2],
                [1,2,0,1,0,2],
                [0,1,0,2,2,3],
                [0,1,1,1,3,1],
                [0,0,4,0,0,1],
                [2,1,1,2,1,4]])
    C = n(P*Q,digits=4)
    pretty_print(C)

    You can manually identify the maximum cost.

    Getting sagemath to identify the largest element is annoyingly unintuitive.

    print(max(C.list()))

    
    
85. 
    

    Question

    prevnext

    In a 15-minute city, all the following grocery stores are within a quick walk, and their prices (in USD) for some foods are shown.

    	apple	banana	cherry	durian	eggplant	fig	grapefruit	honeydew
    Lovely Lunches	1.83	0.91	6.73	7.81	4.23	4.77	3.21	5.05
    More Munchies	1.64	0.85	6.01	9.45	3.77	3.86	3.59	5.81
    Nick’s Necessities	1.73	1.09	5.98	8.48	3.62	3.39	3.34	5.19
    Overpriced Options	1.63	0.96	6.47	8.77	3.90	3.37	2.86	5.17
    Price Pincher	1.30	0.92	6.11	8.11	3.05	4.08	2.34	5.98

    Each of the people below plans to go shopping for those foods. Their grocery lists are shown.

    	Quintin	Rachel	Sebastian	Trevor	Ursula	Venus
    apple	1	0	0	2	0	0
    banana	0	2	0	0	0	2
    cherry	4	4	1	0	1	1
    durian	1	1	0	1	2	0
    eggplant	0	0	4	1	2	4
    fig	3	0	2	0	2	1
    grapefruit	2	2	3	4	3	0
    honeydew	0	1	1	3	4	3

    We can create these matrices in SageMath using the following commands:

    P = matrix([[1.83,0.91,6.73,7.81,4.23,4.77,3.21,5.05],
                [1.64,0.85,6.01,9.45,3.77,3.86,3.59,5.81],
                [1.73,1.09,5.98,8.48,3.62,3.39,3.34,5.19],
                [1.63,0.96,6.47,8.77,3.90,3.37,2.86,5.17],
                [1.30,0.92,6.11,8.11,3.05,4.08,2.34,5.98]])
    
    Q = matrix([[1,0,0,2,0,0],
                [0,2,0,0,0,2],
                [4,4,1,0,1,1],
                [1,1,0,1,2,0],
                [0,0,4,1,2,4],
                [3,0,2,0,2,1],
                [2,2,3,4,3,0],
                [0,1,1,3,4,3]])

    We can find the product of these matrices.

    pretty_print(P*Q)

    What is the maximum possible cost of a grocery trip?

    
    

    Solution

    Show solution

    The maximum cost is 74.18 USD. When you run the lines of code, you should get the following matrix:

    \[\begin{bmatrix} 57.29 & 48.02 & 47.87 & 43.69 & 70.18 & 45.39\\ 53.89 & 48.18 & 45.39 & 48.29 & 74.18 & 44.08\\ 50.98 & 46.45 & 42.45 & 44.49 & 67.74 & 41.6\\ 52.11 & 47.46 & 42.56 & 42.88 & 67.81 & 42.87\\ 50.77 & 45.05 & 39.47 & 41.06 & 67.53 & 42.17 \end{bmatrix}\] If there are many trailing 0s, you can use the n function to get numerical approximations.

    P = matrix([[1.83,0.91,6.73,7.81,4.23,4.77,3.21,5.05],
                [1.64,0.85,6.01,9.45,3.77,3.86,3.59,5.81],
                [1.73,1.09,5.98,8.48,3.62,3.39,3.34,5.19],
                [1.63,0.96,6.47,8.77,3.90,3.37,2.86,5.17],
                [1.30,0.92,6.11,8.11,3.05,4.08,2.34,5.98]])
    Q = matrix([[1,0,0,2,0,0],
                [0,2,0,0,0,2],
                [4,4,1,0,1,1],
                [1,1,0,1,2,0],
                [0,0,4,1,2,4],
                [3,0,2,0,2,1],
                [2,2,3,4,3,0],
                [0,1,1,3,4,3]])
    C = n(P*Q,digits=4)
    pretty_print(C)

    You can manually identify the maximum cost.

    Getting sagemath to identify the largest element is annoyingly unintuitive.

    print(max(C.list()))

    
    
86. 
    

    Question

    prevnext

    In a 15-minute city, all the following grocery stores are within a quick walk, and their prices (in USD) for some foods are shown.

    	apple	banana	cherry	durian	eggplant	fig	grapefruit	honeydew
    Lovely Lunches	1.16	1.02	6.56	8.14	4.03	3.87	3.55	4.57
    More Munchies	1.19	1.03	5.79	9.88	3.44	4.69	2.55	4.94
    Nick’s Necessities	1.73	0.73	6.41	7.02	3.82	4.24	3.38	5.81
    Overpriced Options	1.76	0.96	6.44	9.17	4.59	4.12	3.39	6.61
    Price Pincher	1.28	1.10	5.12	7.13	3.08	4.24	3.57	6.80

    Each of the people below plans to go shopping for those foods. Their grocery lists are shown.

    	Quintin	Rachel	Sebastian	Trevor	Ursula	Venus
    apple	1	2	1	1	0	1
    banana	2	2	0	0	3	2
    cherry	0	4	1	1	4	0
    durian	1	1	0	3	2	0
    eggplant	0	0	2	0	2	2
    fig	2	1	2	2	0	3
    grapefruit	2	3	4	2	1	1
    honeydew	0	1	3	2	1	0

    We can create these matrices in SageMath using the following commands:

    P = matrix([[1.16,1.02,6.56,8.14,4.03,3.87,3.55,4.57],
                [1.19,1.03,5.79,9.88,3.44,4.69,2.55,4.94],
                [1.73,0.73,6.41,7.02,3.82,4.24,3.38,5.81],
                [1.76,0.96,6.44,9.17,4.59,4.12,3.39,6.61],
                [1.28,1.10,5.12,7.13,3.08,4.24,3.57,6.80]])
    
    Q = matrix([[1,2,1,1,0,1],
                [2,2,0,0,3,2],
                [0,4,1,1,4,0],
                [1,1,0,3,2,0],
                [0,0,2,0,2,2],
                [2,1,2,2,0,3],
                [2,3,4,2,1,1],
                [0,1,3,2,1,0]])

    We can find the product of these matrices.

    pretty_print(P*Q)

    What is the maximum possible cost of a grocery trip?

    
    

    Solution

    Show solution

    The maximum cost is 66.16 USD. When you run the lines of code, you should get the following matrix:

    \[\begin{bmatrix} 26.18 & 57.83 & 51.43 & 56.12 & 61.76 & 26.42\\ 27.61 & 54.76 & 48.26 & 60.98 & 60.38 & 26.75\\ 25.45 & 57.77 & 55.21 & 56.06 & 58.7 & 26.93\\ 27.87 & 61.27 & 59.01 & 63.95 & 66.16 & 28.61\\ 26.23 & 54.12 & 55.72 & 57.01 & 54.57 & 25.93 \end{bmatrix}\] If there are many trailing 0s, you can use the n function to get numerical approximations.

    P = matrix([[1.16,1.02,6.56,8.14,4.03,3.87,3.55,4.57],
                [1.19,1.03,5.79,9.88,3.44,4.69,2.55,4.94],
                [1.73,0.73,6.41,7.02,3.82,4.24,3.38,5.81],
                [1.76,0.96,6.44,9.17,4.59,4.12,3.39,6.61],
                [1.28,1.10,5.12,7.13,3.08,4.24,3.57,6.80]])
    Q = matrix([[1,2,1,1,0,1],
                [2,2,0,0,3,2],
                [0,4,1,1,4,0],
                [1,1,0,3,2,0],
                [0,0,2,0,2,2],
                [2,1,2,2,0,3],
                [2,3,4,2,1,1],
                [0,1,3,2,1,0]])
    C = n(P*Q,digits=4)
    pretty_print(C)

    You can manually identify the maximum cost.

    Getting sagemath to identify the largest element is annoyingly unintuitive.

    print(max(C.list()))

    
    
87. 
    

    Question

    prevnext

    In a 15-minute city, all the following grocery stores are within a quick walk, and their prices (in USD) for some foods are shown.

    	apple	banana	cherry	durian	eggplant	fig	grapefruit	honeydew
    Lovely Lunches	1.50	1.00	5.88	9.25	4.73	4.36	3.25	5.65
    More Munchies	1.47	0.83	5.99	9.70	4.87	3.33	2.32	5.83
    Nick’s Necessities	1.66	0.84	4.53	6.54	4.45	3.19	3.43	5.01
    Overpriced Options	1.71	0.95	5.16	9.79	4.16	4.13	3.28	7.23
    Price Pincher	1.48	1.11	6.97	6.65	3.13	3.80	2.53	6.86

    Each of the people below plans to go shopping for those foods. Their grocery lists are shown.

    	Quintin	Rachel	Sebastian	Trevor	Ursula	Venus
    apple	2	1	3	2	0	0
    banana	3	1	2	0	1	3
    cherry	1	3	0	1	2	2
    durian	1	0	0	1	4	2
    eggplant	1	0	1	4	1	4
    fig	4	0	0	2	0	2
    grapefruit	0	2	1	2	2	0
    honeydew	2	2	1	0	1	1

    We can create these matrices in SageMath using the following commands:

    P = matrix([[1.50,1.00,5.88,9.25,4.73,4.36,3.25,5.65],
                [1.47,0.83,5.99,9.70,4.87,3.33,2.32,5.83],
                [1.66,0.84,4.53,6.54,4.45,3.19,3.43,5.01],
                [1.71,0.95,5.16,9.79,4.16,4.13,3.28,7.23],
                [1.48,1.11,6.97,6.65,3.13,3.80,2.53,6.86]])
    
    Q = matrix([[2,1,3,2,0,0],
                [3,1,2,0,1,3],
                [1,3,0,1,2,2],
                [1,0,0,1,4,2],
                [1,0,1,4,1,4],
                [4,0,0,2,0,2],
                [0,2,1,2,2,0],
                [2,2,1,0,1,1]])

    We can find the product of these matrices.

    pretty_print(P*Q)

    What is the maximum possible cost of a grocery trip?

    
    

    Solution

    Show solution

    The maximum cost is 68.38 USD. When you run the lines of code, you should get the following matrix:

    \[\begin{bmatrix} 54.6 & 37.94 & 20.13 & 52.27 & 66.64 & 66.55\\ 50.97 & 36.57 & 19.09 & 49.41 & 66.95 & 65.84\\ 44.14 & 32.97 & 19.55 & 45.43 & 52.38 & 53.85\\ 56.36 & 39.16 & 21.7 & 49.83 & 68.38 & 64.88\\ 51.96 & 42.28 & 19.18 & 41.76 & 56.7 & 57.55 \end{bmatrix}\] If there are many trailing 0s, you can use the n function to get numerical approximations.

    P = matrix([[1.50,1.00,5.88,9.25,4.73,4.36,3.25,5.65],
                [1.47,0.83,5.99,9.70,4.87,3.33,2.32,5.83],
                [1.66,0.84,4.53,6.54,4.45,3.19,3.43,5.01],
                [1.71,0.95,5.16,9.79,4.16,4.13,3.28,7.23],
                [1.48,1.11,6.97,6.65,3.13,3.80,2.53,6.86]])
    Q = matrix([[2,1,3,2,0,0],
                [3,1,2,0,1,3],
                [1,3,0,1,2,2],
                [1,0,0,1,4,2],
                [1,0,1,4,1,4],
                [4,0,0,2,0,2],
                [0,2,1,2,2,0],
                [2,2,1,0,1,1]])
    C = n(P*Q,digits=4)
    pretty_print(C)

    You can manually identify the maximum cost.

    Getting sagemath to identify the largest element is annoyingly unintuitive.

    print(max(C.list()))

    
    
88. 
    

    Question

    prevnext

    In a 15-minute city, all the following grocery stores are within a quick walk, and their prices (in USD) for some foods are shown.

    	apple	banana	cherry	durian	eggplant	fig	grapefruit	honeydew
    Lovely Lunches	1.76	0.74	4.92	6.26	3.39	3.33	2.85	6.93
    More Munchies	1.21	1.01	5.64	9.80	4.12	3.30	2.79	6.96
    Nick’s Necessities	1.55	0.84	4.98	9.90	3.86	4.19	2.33	4.63
    Overpriced Options	1.18	1.05	4.58	6.53	3.62	3.28	3.32	6.99
    Price Pincher	1.30	0.72	7.41	7.01	4.68	4.35	2.97	6.34

    Each of the people below plans to go shopping for those foods. Their grocery lists are shown.

    	Quintin	Rachel	Sebastian	Trevor	Ursula	Venus
    apple	2	3	2	0	1	1
    banana	1	1	0	1	0	2
    cherry	0	0	1	0	0	2
    durian	0	2	3	2	4	0
    eggplant	0	1	1	1	2	0
    fig	0	4	2	4	2	4
    grapefruit	2	1	1	1	2	0
    honeydew	4	0	0	2	1	2

    We can create these matrices in SageMath using the following commands:

    P = matrix([[1.76,0.74,4.92,6.26,3.39,3.33,2.85,6.93],
                [1.21,1.01,5.64,9.80,4.12,3.30,2.79,6.96],
                [1.55,0.84,4.98,9.90,3.86,4.19,2.33,4.63],
                [1.18,1.05,4.58,6.53,3.62,3.28,3.32,6.99],
                [1.30,0.72,7.41,7.01,4.68,4.35,2.97,6.34]])
    
    Q = matrix([[2,3,2,0,1,1],
                [1,1,0,1,0,2],
                [0,0,1,0,0,2],
                [0,2,3,2,4,0],
                [0,1,1,1,2,0],
                [0,4,2,4,2,4],
                [2,1,1,1,2,0],
                [4,0,0,2,1,2]])

    We can find the product of these matrices.

    pretty_print(P*Q)

    What is the maximum possible cost of a grocery trip?

    
    

    Solution

    Show solution

    The maximum cost is 67.79 USD. When you run the lines of code, you should get the following matrix:

    \[\begin{bmatrix} 37.68 & 38.1 & 40.12 & 46.68 & 52.87 & 40.26\\ 36.85 & 44.35 & 50.97 & 54.64 & 67.79 & 41.63\\ 27.12 & 48.24 & 52.35 & 52.85 & 66.54 & 39.21\\ 38.01 & 37.71 & 40.03 & 48.15 & 54.73 & 39.54\\ 34.62 & 43.69 & 47.39 & 52.47 & 59.68 & 47.64 \end{bmatrix}\] If there are many trailing 0s, you can use the n function to get numerical approximations.

    P = matrix([[1.76,0.74,4.92,6.26,3.39,3.33,2.85,6.93],
                [1.21,1.01,5.64,9.80,4.12,3.30,2.79,6.96],
                [1.55,0.84,4.98,9.90,3.86,4.19,2.33,4.63],
                [1.18,1.05,4.58,6.53,3.62,3.28,3.32,6.99],
                [1.30,0.72,7.41,7.01,4.68,4.35,2.97,6.34]])
    Q = matrix([[2,3,2,0,1,1],
                [1,1,0,1,0,2],
                [0,0,1,0,0,2],
                [0,2,3,2,4,0],
                [0,1,1,1,2,0],
                [0,4,2,4,2,4],
                [2,1,1,1,2,0],
                [4,0,0,2,1,2]])
    C = n(P*Q,digits=4)
    pretty_print(C)

    You can manually identify the maximum cost.

    Getting sagemath to identify the largest element is annoyingly unintuitive.

    print(max(C.list()))

    
    
89. 
    

    Question

    prevnext

    In a 15-minute city, all the following grocery stores are within a quick walk, and their prices (in USD) for some foods are shown.

    	apple	banana	cherry	durian	eggplant	fig	grapefruit	honeydew
    Lovely Lunches	1.21	1.06	7.37	6.20	3.99	4.66	3.66	6.21
    More Munchies	1.60	0.76	4.83	7.77	3.48	4.16	3.16	5.83
    Nick’s Necessities	1.26	0.95	6.74	7.47	4.07	3.84	3.05	6.84
    Overpriced Options	1.55	1.11	6.58	6.82	4.54	3.84	2.38	6.80
    Price Pincher	1.31	0.94	6.08	9.14	3.14	3.10	3.60	6.21

    Each of the people below plans to go shopping for those foods. Their grocery lists are shown.

    	Quintin	Rachel	Sebastian	Trevor	Ursula	Venus
    apple	3	2	0	3	0	0
    banana	2	0	1	2	1	2
    cherry	0	1	4	0	4	1
    durian	4	2	1	1	2	2
    eggplant	1	2	0	0	1	0
    fig	2	1	2	0	2	2
    grapefruit	0	4	0	0	3	0
    honeydew	0	1	0	1	2	3

    We can create these matrices in SageMath using the following commands:

    P = matrix([[1.21,1.06,7.37,6.20,3.99,4.66,3.66,6.21],
                [1.60,0.76,4.83,7.77,3.48,4.16,3.16,5.83],
                [1.26,0.95,6.74,7.47,4.07,3.84,3.05,6.84],
                [1.55,1.11,6.58,6.82,4.54,3.84,2.38,6.80],
                [1.31,0.94,6.08,9.14,3.14,3.10,3.60,6.21]])
    
    Q = matrix([[3,2,0,3,0,0],
                [2,0,1,2,1,2],
                [0,1,4,0,4,1],
                [4,2,1,1,2,2],
                [1,2,0,0,1,0],
                [2,1,2,0,2,2],
                [0,4,0,0,3,0],
                [0,1,0,1,2,3]])

    We can find the product of these matrices.

    pretty_print(P*Q)

    What is the maximum possible cost of a grocery trip?

    
    

    Solution

    Show solution

    The maximum cost is 79.65 USD. When you run the lines of code, you should get the following matrix:

    \[\begin{bmatrix} 43.86 & 55.68 & 46.06 & 18.16 & 79.65 & 49.84\\ 49.2 & 53.16 & 36.17 & 19.92 & 68.56 & 47.7\\ 47.31 & 55.22 & 43.06 & 19.99 & 77.43 & 51.78\\ 46.37 & 52.56 & 41.93 & 20.49 & 74.03 & 50.52\\ 51.71 & 56.97 & 40.6 & 21.16 & 76.1 & 51.07 \end{bmatrix}\] If there are many trailing 0s, you can use the n function to get numerical approximations.

    P = matrix([[1.21,1.06,7.37,6.20,3.99,4.66,3.66,6.21],
                [1.60,0.76,4.83,7.77,3.48,4.16,3.16,5.83],
                [1.26,0.95,6.74,7.47,4.07,3.84,3.05,6.84],
                [1.55,1.11,6.58,6.82,4.54,3.84,2.38,6.80],
                [1.31,0.94,6.08,9.14,3.14,3.10,3.60,6.21]])
    Q = matrix([[3,2,0,3,0,0],
                [2,0,1,2,1,2],
                [0,1,4,0,4,1],
                [4,2,1,1,2,2],
                [1,2,0,0,1,0],
                [2,1,2,0,2,2],
                [0,4,0,0,3,0],
                [0,1,0,1,2,3]])
    C = n(P*Q,digits=4)
    pretty_print(C)

    You can manually identify the maximum cost.

    Getting sagemath to identify the largest element is annoyingly unintuitive.

    print(max(C.list()))

    
    
90. 
    

    Question

    prevnext

    In a 15-minute city, all the following grocery stores are within a quick walk, and their prices (in USD) for some foods are shown.

    	apple	banana	cherry	durian	eggplant	fig	grapefruit	honeydew
    Lovely Lunches	1.85	0.86	7.01	8.26	3.23	3.38	2.68	5.47
    More Munchies	1.48	1.04	6.28	7.05	4.61	3.12	3.59	6.74
    Nick’s Necessities	1.49	0.96	7.00	9.95	3.53	4.87	3.72	5.52
    Overpriced Options	1.51	0.90	4.67	8.91	4.85	4.99	2.40	5.75
    Price Pincher	1.59	0.77	5.23	7.43	3.42	4.26	3.64	6.08

    Each of the people below plans to go shopping for those foods. Their grocery lists are shown.

    	Quintin	Rachel	Sebastian	Trevor	Ursula	Venus
    apple	2	2	1	1	0	2
    banana	1	1	2	2	1	1
    cherry	0	4	0	2	1	3
    durian	0	2	0	1	1	4
    eggplant	3	0	0	0	3	2
    fig	2	0	2	0	4	1
    grapefruit	0	1	2	4	0	0
    honeydew	1	0	1	3	0	0

    We can create these matrices in SageMath using the following commands:

    P = matrix([[1.85,0.86,7.01,8.26,3.23,3.38,2.68,5.47],
                [1.48,1.04,6.28,7.05,4.61,3.12,3.59,6.74],
                [1.49,0.96,7.00,9.95,3.53,4.87,3.72,5.52],
                [1.51,0.90,4.67,8.91,4.85,4.99,2.40,5.75],
                [1.59,0.77,5.23,7.43,3.42,4.26,3.64,6.08]])
    
    Q = matrix([[2,2,1,1,0,2],
                [1,1,2,2,1,1],
                [0,4,0,2,1,3],
                [0,2,0,1,1,4],
                [3,0,0,0,3,2],
                [2,0,2,0,4,1],
                [0,1,2,4,0,0],
                [1,0,1,3,0,0]])

    We can find the product of these matrices.

    pretty_print(P*Q)

    What is the maximum possible cost of a grocery trip?

    
    

    Solution

    Show solution

    The maximum cost is 76.67 USD. When you run the lines of code, you should get the following matrix:

    \[\begin{bmatrix} 26.48 & 51.8 & 21.16 & 52.98 & 39.34 & 68.47\\ 30.81 & 46.81 & 23.72 & 57.75 & 40.68 & 63.38\\ 29.79 & 55.56 & 26.11 & 58.8 & 47.98 & 76.67\\ 34.2 & 42.82 & 23.84 & 48.41 & 48.99 & 68.26\\ 28.81 & 43.37 & 25.01 & 53.82 & 40.73 & 60.46 \end{bmatrix}\] If there are many trailing 0s, you can use the n function to get numerical approximations.

    P = matrix([[1.85,0.86,7.01,8.26,3.23,3.38,2.68,5.47],
                [1.48,1.04,6.28,7.05,4.61,3.12,3.59,6.74],
                [1.49,0.96,7.00,9.95,3.53,4.87,3.72,5.52],
                [1.51,0.90,4.67,8.91,4.85,4.99,2.40,5.75],
                [1.59,0.77,5.23,7.43,3.42,4.26,3.64,6.08]])
    Q = matrix([[2,2,1,1,0,2],
                [1,1,2,2,1,1],
                [0,4,0,2,1,3],
                [0,2,0,1,1,4],
                [3,0,0,0,3,2],
                [2,0,2,0,4,1],
                [0,1,2,4,0,0],
                [1,0,1,3,0,0]])
    C = n(P*Q,digits=4)
    pretty_print(C)

    You can manually identify the maximum cost.

    Getting sagemath to identify the largest element is annoyingly unintuitive.

    print(max(C.list()))

    
    
91. 
    

    Question

    prevnext

    Let matrix \(M\) be defined below:

    \[M = \begin{bmatrix} 3 & 5 & 4\\ 1 & 2 & 6 \end{bmatrix}\]

    Which matrix equals \(M^\mathrm{T}\), the transpose of \(M\)?

    \[A = \begin{bmatrix} 3 & 5 & 4\\ 1 & 2 & 6 \end{bmatrix}\] \[B = \begin{bmatrix} 3 & 1 & 5\\ 2 & 4 & 6 \end{bmatrix}\] \[C = \begin{bmatrix} 3 & 2\\ 1 & 4\\ 5 & 6 \end{bmatrix}\] \[D = \begin{bmatrix} 3 & 1\\ 5 & 2\\ 4 & 6 \end{bmatrix}\]

    
    
    1. A
    2. B
    3. C
    4. D
    
    

    Solution

    Show solution

    You can use SageMath.

    M = matrix([[3,5,4],
                [1,2,6]])
    print(M.transpose())

    
    
92. 
    

    Question

    prevnext

    Let matrix \(M\) be defined below:

    \[M = \begin{bmatrix} 9 & 10 & 12\\ 2 & 8 & 6\\ 4 & 5 & 1\\ 3 & 11 & 7 \end{bmatrix}\]

    Which matrix equals \(M^\mathrm{T}\), the transpose of \(M\)?

    \[A = \begin{bmatrix} 9 & 2 & 4\\ 3 & 10 & 8\\ 5 & 11 & 12\\ 6 & 1 & 7 \end{bmatrix}\] \[B = \begin{bmatrix} 9 & 3 & 5 & 6\\ 2 & 10 & 11 & 1\\ 4 & 8 & 12 & 7 \end{bmatrix}\] \[C = \begin{bmatrix} 9 & 10 & 12\\ 2 & 8 & 6\\ 4 & 5 & 1\\ 3 & 11 & 7 \end{bmatrix}\] \[D = \begin{bmatrix} 9 & 2 & 4 & 3\\ 10 & 8 & 5 & 11\\ 12 & 6 & 1 & 7 \end{bmatrix}\]

    
    
    1. A
    2. B
    3. C
    4. D
    
    

    Solution

    Show solution

    You can use SageMath.

    M = matrix([[9,10,12],
                [2,8,6],
                [4,5,1],
                [3,11,7]])
    print(M.transpose())

    
    
93. 
    

    Question

    prevnext

    Let matrix \(M\) be defined below:

    \[M = \begin{bmatrix} 1 & 4 & 3\\ 2 & 6 & 5 \end{bmatrix}\]

    Which matrix equals \(M^\mathrm{T}\), the transpose of \(M\)?

    \[A = \begin{bmatrix} 1 & 4 & 3\\ 2 & 6 & 5 \end{bmatrix}\] \[B = \begin{bmatrix} 1 & 2 & 4\\ 6 & 3 & 5 \end{bmatrix}\] \[C = \begin{bmatrix} 1 & 6\\ 2 & 3\\ 4 & 5 \end{bmatrix}\] \[D = \begin{bmatrix} 1 & 2\\ 4 & 6\\ 3 & 5 \end{bmatrix}\]

    
    
    1. A
    2. B
    3. C
    4. D
    
    

    Solution

    Show solution

    You can use SageMath.

    M = matrix([[1,4,3],
                [2,6,5]])
    print(M.transpose())

    
    
94. 
    

    Question

    prevnext

    Let matrix \(M\) be defined below:

    \[M = \begin{bmatrix} 2 & 11 & 12 & 6\\ 4 & 1 & 5 & 3\\ 8 & 7 & 10 & 9 \end{bmatrix}\]

    Which matrix equals \(M^\mathrm{T}\), the transpose of \(M\)?

    \[A = \begin{bmatrix} 2 & 11 & 12 & 6\\ 4 & 1 & 5 & 3\\ 8 & 7 & 10 & 9 \end{bmatrix}\] \[B = \begin{bmatrix} 2 & 1 & 10\\ 4 & 7 & 6\\ 8 & 12 & 3\\ 11 & 5 & 9 \end{bmatrix}\] \[C = \begin{bmatrix} 2 & 4 & 8 & 11\\ 1 & 7 & 12 & 5\\ 10 & 6 & 3 & 9 \end{bmatrix}\] \[D = \begin{bmatrix} 2 & 4 & 8\\ 11 & 1 & 7\\ 12 & 5 & 10\\ 6 & 3 & 9 \end{bmatrix}\]

    
    
    1. A
    2. B
    3. C
    4. D
    
    

    Solution

    Show solution

    You can use SageMath.

    M = matrix([[2,11,12,6],
                [4,1,5,3],
                [8,7,10,9]])
    print(M.transpose())

    
    
95. 
    

    Question

    prevnext

    Let matrix \(M\) be defined below:

    \[M = \begin{bmatrix} 6 & 8\\ 4 & 5\\ 2 & 3\\ 7 & 1 \end{bmatrix}\]

    Which matrix equals \(M^\mathrm{T}\), the transpose of \(M\)?

    \[A = \begin{bmatrix} 6 & 4 & 2 & 7\\ 8 & 5 & 3 & 1 \end{bmatrix}\] \[B = \begin{bmatrix} 6 & 4\\ 2 & 7\\ 8 & 5\\ 3 & 1 \end{bmatrix}\] \[C = \begin{bmatrix} 6 & 2 & 8 & 3\\ 4 & 7 & 5 & 1 \end{bmatrix}\] \[D = \begin{bmatrix} 6 & 8\\ 4 & 5\\ 2 & 3\\ 7 & 1 \end{bmatrix}\]

    
    
    1. A
    2. B
    3. C
    4. D
    
    

    Solution

    Show solution

    You can use SageMath.

    M = matrix([[6,8],
                [4,5],
                [2,3],
                [7,1]])
    print(M.transpose())

    
    
96. 
    

    Question

    prevnext

    Let matrix \(M\) be defined below:

    \[M = \begin{bmatrix} 3 & 6\\ 1 & 2\\ 5 & 4 \end{bmatrix}\]

    Which matrix equals \(M^\mathrm{T}\), the transpose of \(M\)?

    \[A = \begin{bmatrix} 3 & 6\\ 1 & 2\\ 5 & 4 \end{bmatrix}\] \[B = \begin{bmatrix} 3 & 5 & 2\\ 1 & 6 & 4 \end{bmatrix}\] \[C = \begin{bmatrix} 3 & 1\\ 5 & 6\\ 2 & 4 \end{bmatrix}\] \[D = \begin{bmatrix} 3 & 1 & 5\\ 6 & 2 & 4 \end{bmatrix}\]

    
    
    1. A
    2. B
    3. C
    4. D
    
    

    Solution

    Show solution

    You can use SageMath.

    M = matrix([[3,6],
                [1,2],
                [5,4]])
    print(M.transpose())

    
    
97. 
    

    Question

    prevnext

    Let matrix \(M\) be defined below:

    \[M = \begin{bmatrix} 5 & 4\\ 2 & 1\\ 6 & 3 \end{bmatrix}\]

    Which matrix equals \(M^\mathrm{T}\), the transpose of \(M\)?

    \[A = \begin{bmatrix} 5 & 2 & 6\\ 4 & 1 & 3 \end{bmatrix}\] \[B = \begin{bmatrix} 5 & 6 & 1\\ 2 & 4 & 3 \end{bmatrix}\] \[C = \begin{bmatrix} 5 & 2\\ 6 & 4\\ 1 & 3 \end{bmatrix}\] \[D = \begin{bmatrix} 5 & 4\\ 2 & 1\\ 6 & 3 \end{bmatrix}\]

    
    
    1. A
    2. B
    3. C
    4. D
    
    

    Solution

    Show solution

    You can use SageMath.

    M = matrix([[5,4],
                [2,1],
                [6,3]])
    print(M.transpose())

    
    
98. 
    

    Question

    prevnext

    Let matrix \(M\) be defined below:

    \[M = \begin{bmatrix} 1 & 3\\ 5 & 8\\ 2 & 7\\ 4 & 6 \end{bmatrix}\]

    Which matrix equals \(M^\mathrm{T}\), the transpose of \(M\)?

    \[A = \begin{bmatrix} 1 & 5\\ 2 & 4\\ 3 & 8\\ 7 & 6 \end{bmatrix}\] \[B = \begin{bmatrix} 1 & 5 & 2 & 4\\ 3 & 8 & 7 & 6 \end{bmatrix}\] \[C = \begin{bmatrix} 1 & 3\\ 5 & 8\\ 2 & 7\\ 4 & 6 \end{bmatrix}\] \[D = \begin{bmatrix} 1 & 2 & 3 & 7\\ 5 & 4 & 8 & 6 \end{bmatrix}\]

    
    
    1. A
    2. B
    3. C
    4. D
    
    

    Solution

    Show solution

    You can use SageMath.

    M = matrix([[1,3],
                [5,8],
                [2,7],
                [4,6]])
    print(M.transpose())

    
    
99. 
    

    Question

    prevnext

    Let matrix \(M\) be defined below:

    \[M = \begin{bmatrix} 2 & 1\\ 4 & 5\\ 3 & 6 \end{bmatrix}\]

    Which matrix equals \(M^\mathrm{T}\), the transpose of \(M\)?

    \[A = \begin{bmatrix} 2 & 4 & 3\\ 1 & 5 & 6 \end{bmatrix}\] \[B = \begin{bmatrix} 2 & 4\\ 3 & 1\\ 5 & 6 \end{bmatrix}\] \[C = \begin{bmatrix} 2 & 3 & 5\\ 4 & 1 & 6 \end{bmatrix}\] \[D = \begin{bmatrix} 2 & 1\\ 4 & 5\\ 3 & 6 \end{bmatrix}\]

    
    
    1. A
    2. B
    3. C
    4. D
    
    

    Solution

    Show solution

    You can use SageMath.

    M = matrix([[2,1],
                [4,5],
                [3,6]])
    print(M.transpose())

    
    
100. 
     

     Question

     prevnext

     Let matrix \(M\) be defined below:

     \[M = \begin{bmatrix} 5 & 2\\ 1 & 8\\ 3 & 7\\ 4 & 6 \end{bmatrix}\]

     Which matrix equals \(M^\mathrm{T}\), the transpose of \(M\)?

     \[A = \begin{bmatrix} 5 & 1 & 3 & 4\\ 2 & 8 & 7 & 6 \end{bmatrix}\] \[B = \begin{bmatrix} 5 & 1\\ 3 & 4\\ 2 & 8\\ 7 & 6 \end{bmatrix}\] \[C = \begin{bmatrix} 5 & 2\\ 1 & 8\\ 3 & 7\\ 4 & 6 \end{bmatrix}\] \[D = \begin{bmatrix} 5 & 3 & 2 & 7\\ 1 & 4 & 8 & 6 \end{bmatrix}\]

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     You can use SageMath.

     M = matrix([[5,2],
                 [1,8],
                 [3,7],
                 [4,6]])
     print(M.transpose())

     
     
101. 
     

     Question

     prevnext

     The 2D rotation matrix is defined below for a counter-clockwise rotation by an angle \(\theta\) when it is the first operand in matrix multiplication.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \]

     Let \(\theta=3.61\) radians. Determine \(R\) (approximate decimal values).

     \(R =\)

     -------- -0.89-0.450.450.89	-------- -0.89-0.450.450.89
     -------- -0.89-0.450.450.89	-------- -0.89-0.450.450.89

     
     
     
     

     Solution

     Show solution

     \[R = \begin{bmatrix} -0.89 & 0.45\\ -0.45 & -0.89 \end{bmatrix}\]

     
     
102. 
     

     Question

     prevnext

     The 2D rotation matrix is defined below for a counter-clockwise rotation by an angle \(\theta\) when it is the first operand in matrix multiplication.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \]

     Let \(\theta=5.60\) radians. Determine \(R\) (approximate decimal values).

     \(R =\)

     -------- -0.78-0.630.630.78	-------- -0.78-0.630.630.78
     -------- -0.78-0.630.630.78	-------- -0.78-0.630.630.78

     
     
     
     

     Solution

     Show solution

     \[R = \begin{bmatrix} 0.78 & 0.63\\ -0.63 & 0.78 \end{bmatrix}\]

     
     
103. 
     

     Question

     prevnext

     The 2D rotation matrix is defined below for a counter-clockwise rotation by an angle \(\theta\) when it is the first operand in matrix multiplication.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \]

     Let \(\theta=2.21\) radians. Determine \(R\) (approximate decimal values).

     \(R =\)

     -------- -0.8-0.60.60.8	-------- -0.8-0.60.60.8
     -------- -0.8-0.60.60.8	-------- -0.8-0.60.60.8

     
     
     
     

     Solution

     Show solution

     \[R = \begin{bmatrix} -0.6 & -0.8\\ 0.8 & -0.6 \end{bmatrix}\]

     
     
104. 
     

     Question

     prevnext

     The 2D rotation matrix is defined below for a counter-clockwise rotation by an angle \(\theta\) when it is the first operand in matrix multiplication.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \]

     Let \(\theta=5.91\) radians. Determine \(R\) (approximate decimal values).

     \(R =\)

     -------- -0.93-0.360.360.93	-------- -0.93-0.360.360.93
     -------- -0.93-0.360.360.93	-------- -0.93-0.360.360.93

     
     
     
     

     Solution

     Show solution

     \[R = \begin{bmatrix} 0.93 & 0.36\\ -0.36 & 0.93 \end{bmatrix}\]

     
     
105. 
     

     Question

     prevnext

     The 2D rotation matrix is defined below for a counter-clockwise rotation by an angle \(\theta\) when it is the first operand in matrix multiplication.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \]

     Let \(\theta=2.24\) radians. Determine \(R\) (approximate decimal values).

     \(R =\)

     -------- -0.78-0.620.620.78	-------- -0.78-0.620.620.78
     -------- -0.78-0.620.620.78	-------- -0.78-0.620.620.78

     
     
     
     

     Solution

     Show solution

     \[R = \begin{bmatrix} -0.62 & -0.78\\ 0.78 & -0.62 \end{bmatrix}\]

     
     
106. 
     

     Question

     prevnext

     The 2D rotation matrix is defined below for a counter-clockwise rotation by an angle \(\theta\) when it is the first operand in matrix multiplication.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \]

     Let \(\theta=5.99\) radians. Determine \(R\) (approximate decimal values).

     \(R =\)

     -------- -0.96-0.290.290.96	-------- -0.96-0.290.290.96
     -------- -0.96-0.290.290.96	-------- -0.96-0.290.290.96

     
     
     
     

     Solution

     Show solution

     \[R = \begin{bmatrix} 0.96 & 0.29\\ -0.29 & 0.96 \end{bmatrix}\]

     
     
107. 
     

     Question

     prevnext

     The 2D rotation matrix is defined below for a counter-clockwise rotation by an angle \(\theta\) when it is the first operand in matrix multiplication.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \]

     Let \(\theta=2.48\) radians. Determine \(R\) (approximate decimal values).

     \(R =\)

     -------- -0.79-0.610.610.79	-------- -0.79-0.610.610.79
     -------- -0.79-0.610.610.79	-------- -0.79-0.610.610.79

     
     
     
     

     Solution

     Show solution

     \[R = \begin{bmatrix} -0.79 & -0.61\\ 0.61 & -0.79 \end{bmatrix}\]

     
     
108. 
     

     Question

     prevnext

     The 2D rotation matrix is defined below for a counter-clockwise rotation by an angle \(\theta\) when it is the first operand in matrix multiplication.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \]

     Let \(\theta=4.41\) radians. Determine \(R\) (approximate decimal values).

     \(R =\)

     -------- -0.95-0.30.30.95	-------- -0.95-0.30.30.95
     -------- -0.95-0.30.30.95	-------- -0.95-0.30.30.95

     
     
     
     

     Solution

     Show solution

     \[R = \begin{bmatrix} -0.3 & 0.95\\ -0.95 & -0.3 \end{bmatrix}\]

     
     
109. 
     

     Question

     prevnext

     The 2D rotation matrix is defined below for a counter-clockwise rotation by an angle \(\theta\) when it is the first operand in matrix multiplication.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \]

     Let \(\theta=0.63\) radians. Determine \(R\) (approximate decimal values).

     \(R =\)

     -------- -0.81-0.590.590.81	-------- -0.81-0.590.590.81
     -------- -0.81-0.590.590.81	-------- -0.81-0.590.590.81

     
     
     
     

     Solution

     Show solution

     \[R = \begin{bmatrix} 0.81 & -0.59\\ 0.59 & 0.81 \end{bmatrix}\]

     
     
110. 
     

     Question

     prevnext

     The 2D rotation matrix is defined below for a counter-clockwise rotation by an angle \(\theta\) when it is the first operand in matrix multiplication.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \]

     Let \(\theta=4.97\) radians. Determine \(R\) (approximate decimal values).

     \(R =\)

     -------- -0.97-0.250.250.97	-------- -0.97-0.250.250.97
     -------- -0.97-0.250.250.97	-------- -0.97-0.250.250.97

     
     
     
     

     Solution

     Show solution

     \[R = \begin{bmatrix} 0.25 & 0.97\\ -0.97 & 0.25 \end{bmatrix}\]

     
     
111. 
     

     Question

     prevnext

     Matrix of original points (preimage)

     A list of 2D-cartesian points is stored as the preimage matrix. Each column represents one point.

     \[P = \begin{bmatrix} 4 & 3 & 7 & 5 & -4 & 1 & -6 & -6 & 2 & -2 & -6\\ 4 & -2 & -1 & 5 & -1 & -2 & 7 & 2 & 4 & 7 & -5 \end{bmatrix}\]

     P = matrix([[4,3,7,5,-4,1,-6,-6,2,-2,-6],[4,-2,-1,5,-1,-2,7,2,4,7,-5]])

     Rotation matrix

     We want to rotate the points by \(\theta=0.93\) radians counterclockwise. We will do this by matrix multiplication with the 2D rotation matrix.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\]

     Image

     We can apply the rotation to the points through matrix multiplication. Remember, matrix multiplication is not commutative. Let \(T\) represent a matrix holding the final positions of the image (after the transformation).

     \[T = RP\]

     After you calculate \(T\), you can plot the points in \(T\) with the following code (in SageMath):

     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     Which is the graph of the points in \(T\)? (Please excuse minor stylistic differences due to my graphs being made in R, not SageMath.)

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     P = matrix([[4,3,7,5,-4,1,-6,-6,2,-2,-6],[4,-2,-1,5,-1,-2,7,2,4,7,-5]])
     theta = 0.93
     R = matrix([[cos(theta),-sin(theta)],
                 [sin(theta),cos(theta)]])
     T = R*P
     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     
     
112. 
     

     Question

     prevnext

     Matrix of original points (preimage)

     A list of 2D-cartesian points is stored as the preimage matrix. Each column represents one point.

     \[P = \begin{bmatrix} 4 & 6 & 1 & -6 & 3 & 3 & 5 & 5 & -5 & 3 & -4 & -2\\ -3 & 2 & 6 & 4 & 2 & -6 & -1 & 3 & 1 & -6 & -3 & 5 \end{bmatrix}\]

     P = matrix([[4,6,1,-6,3,3,5,5,-5,3,-4,-2],[-3,2,6,4,2,-6,-1,3,1,-6,-3,5]])

     Rotation matrix

     We want to rotate the points by \(\theta=4.44\) radians counterclockwise. We will do this by matrix multiplication with the 2D rotation matrix.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\]

     Image

     We can apply the rotation to the points through matrix multiplication. Remember, matrix multiplication is not commutative. Let \(T\) represent a matrix holding the final positions of the image (after the transformation).

     \[T = RP\]

     After you calculate \(T\), you can plot the points in \(T\) with the following code (in SageMath):

     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     Which is the graph of the points in \(T\)? (Please excuse minor stylistic differences due to my graphs being made in R, not SageMath.)

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     P = matrix([[4,6,1,-6,3,3,5,5,-5,3,-4,-2],[-3,2,6,4,2,-6,-1,3,1,-6,-3,5]])
     theta = 4.44
     R = matrix([[cos(theta),-sin(theta)],
                 [sin(theta),cos(theta)]])
     T = R*P
     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     
     
113. 
     

     Question

     prevnext

     Matrix of original points (preimage)

     A list of 2D-cartesian points is stored as the preimage matrix. Each column represents one point.

     \[P = \begin{bmatrix} -7 & -3 & -5 & 4 & -1 & 7 & 3 & -7 & -5 & 4 & -6\\ -3 & 4 & -5 & 7 & -1 & 2 & -6 & -1 & -5 & 6 & 0 \end{bmatrix}\]

     P = matrix([[-7,-3,-5,4,-1,7,3,-7,-5,4,-6],[-3,4,-5,7,-1,2,-6,-1,-5,6,0]])

     Rotation matrix

     We want to rotate the points by \(\theta=4.98\) radians counterclockwise. We will do this by matrix multiplication with the 2D rotation matrix.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\]

     Image

     We can apply the rotation to the points through matrix multiplication. Remember, matrix multiplication is not commutative. Let \(T\) represent a matrix holding the final positions of the image (after the transformation).

     \[T = RP\]

     After you calculate \(T\), you can plot the points in \(T\) with the following code (in SageMath):

     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     Which is the graph of the points in \(T\)? (Please excuse minor stylistic differences due to my graphs being made in R, not SageMath.)

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     P = matrix([[-7,-3,-5,4,-1,7,3,-7,-5,4,-6],[-3,4,-5,7,-1,2,-6,-1,-5,6,0]])
     theta = 4.98
     R = matrix([[cos(theta),-sin(theta)],
                 [sin(theta),cos(theta)]])
     T = R*P
     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     
     
114. 
     

     Question

     prevnext

     Matrix of original points (preimage)

     A list of 2D-cartesian points is stored as the preimage matrix. Each column represents one point.

     \[P = \begin{bmatrix} 3 & 5 & 3 & 3 & -5 & 1 & 5 & -4 & 1 & 7 & 6 & 2\\ -1 & 0 & 7 & 1 & 0 & 1 & -7 & -7 & -1 & -6 & 0 & -1 \end{bmatrix}\]

     P = matrix([[3,5,3,3,-5,1,5,-4,1,7,6,2],[-1,0,7,1,0,1,-7,-7,-1,-6,0,-1]])

     Rotation matrix

     We want to rotate the points by \(\theta=5.37\) radians counterclockwise. We will do this by matrix multiplication with the 2D rotation matrix.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\]

     Image

     We can apply the rotation to the points through matrix multiplication. Remember, matrix multiplication is not commutative. Let \(T\) represent a matrix holding the final positions of the image (after the transformation).

     \[T = RP\]

     After you calculate \(T\), you can plot the points in \(T\) with the following code (in SageMath):

     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     Which is the graph of the points in \(T\)? (Please excuse minor stylistic differences due to my graphs being made in R, not SageMath.)

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     P = matrix([[3,5,3,3,-5,1,5,-4,1,7,6,2],[-1,0,7,1,0,1,-7,-7,-1,-6,0,-1]])
     theta = 5.37
     R = matrix([[cos(theta),-sin(theta)],
                 [sin(theta),cos(theta)]])
     T = R*P
     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     
     
115. 
     

     Question

     prevnext

     Matrix of original points (preimage)

     A list of 2D-cartesian points is stored as the preimage matrix. Each column represents one point.

     \[P = \begin{bmatrix} -5 & -7 & 7 & -7 & 6 & 2 & 6 & -6 & -2 & 6 & -5\\ -2 & 0 & -6 & -1 & 6 & 2 & -7 & 1 & -1 & -7 & 5 \end{bmatrix}\]

     P = matrix([[-5,-7,7,-7,6,2,6,-6,-2,6,-5],[-2,0,-6,-1,6,2,-7,1,-1,-7,5]])

     Rotation matrix

     We want to rotate the points by \(\theta=3.36\) radians counterclockwise. We will do this by matrix multiplication with the 2D rotation matrix.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\]

     Image

     We can apply the rotation to the points through matrix multiplication. Remember, matrix multiplication is not commutative. Let \(T\) represent a matrix holding the final positions of the image (after the transformation).

     \[T = RP\]

     After you calculate \(T\), you can plot the points in \(T\) with the following code (in SageMath):

     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     Which is the graph of the points in \(T\)? (Please excuse minor stylistic differences due to my graphs being made in R, not SageMath.)

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     P = matrix([[-5,-7,7,-7,6,2,6,-6,-2,6,-5],[-2,0,-6,-1,6,2,-7,1,-1,-7,5]])
     theta = 3.36
     R = matrix([[cos(theta),-sin(theta)],
                 [sin(theta),cos(theta)]])
     T = R*P
     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     
     
116. 
     

     Question

     prevnext

     Matrix of original points (preimage)

     A list of 2D-cartesian points is stored as the preimage matrix. Each column represents one point.

     \[P = \begin{bmatrix} -2 & -4 & -4 & -7 & -4 & 7 & -5 & 7 & 7\\ -1 & -2 & 6 & 2 & 7 & -6 & -4 & -6 & 7 \end{bmatrix}\]

     P = matrix([[-2,-4,-4,-7,-4,7,-5,7,7],[-1,-2,6,2,7,-6,-4,-6,7]])

     Rotation matrix

     We want to rotate the points by \(\theta=0.71\) radians counterclockwise. We will do this by matrix multiplication with the 2D rotation matrix.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\]

     Image

     We can apply the rotation to the points through matrix multiplication. Remember, matrix multiplication is not commutative. Let \(T\) represent a matrix holding the final positions of the image (after the transformation).

     \[T = RP\]

     After you calculate \(T\), you can plot the points in \(T\) with the following code (in SageMath):

     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     Which is the graph of the points in \(T\)? (Please excuse minor stylistic differences due to my graphs being made in R, not SageMath.)

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     P = matrix([[-2,-4,-4,-7,-4,7,-5,7,7],[-1,-2,6,2,7,-6,-4,-6,7]])
     theta = 0.71
     R = matrix([[cos(theta),-sin(theta)],
                 [sin(theta),cos(theta)]])
     T = R*P
     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     
     
117. 
     

     Question

     prevnext

     Matrix of original points (preimage)

     A list of 2D-cartesian points is stored as the preimage matrix. Each column represents one point.

     \[P = \begin{bmatrix} -2 & 2 & 6 & 2 & -7 & 5 & 5 & -2 & 5\\ -7 & 5 & 4 & -6 & 6 & 6 & -5 & 6 & -3 \end{bmatrix}\]

     P = matrix([[-2,2,6,2,-7,5,5,-2,5],[-7,5,4,-6,6,6,-5,6,-3]])

     Rotation matrix

     We want to rotate the points by \(\theta=4.63\) radians counterclockwise. We will do this by matrix multiplication with the 2D rotation matrix.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\]

     Image

     We can apply the rotation to the points through matrix multiplication. Remember, matrix multiplication is not commutative. Let \(T\) represent a matrix holding the final positions of the image (after the transformation).

     \[T = RP\]

     After you calculate \(T\), you can plot the points in \(T\) with the following code (in SageMath):

     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     Which is the graph of the points in \(T\)? (Please excuse minor stylistic differences due to my graphs being made in R, not SageMath.)

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     P = matrix([[-2,2,6,2,-7,5,5,-2,5],[-7,5,4,-6,6,6,-5,6,-3]])
     theta = 4.63
     R = matrix([[cos(theta),-sin(theta)],
                 [sin(theta),cos(theta)]])
     T = R*P
     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     
     
118. 
     

     Question

     prevnext

     Matrix of original points (preimage)

     A list of 2D-cartesian points is stored as the preimage matrix. Each column represents one point.

     \[P = \begin{bmatrix} -2 & -1 & 0 & 4 & -4 & -6 & -3 & 2 & 6 & 7 & -1 & -1\\ 5 & -2 & -7 & 7 & 6 & 2 & -7 & -4 & -1 & -3 & 6 & -6 \end{bmatrix}\]

     P = matrix([[-2,-1,0,4,-4,-6,-3,2,6,7,-1,-1],[5,-2,-7,7,6,2,-7,-4,-1,-3,6,-6]])

     Rotation matrix

     We want to rotate the points by \(\theta=1.42\) radians counterclockwise. We will do this by matrix multiplication with the 2D rotation matrix.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\]

     Image

     We can apply the rotation to the points through matrix multiplication. Remember, matrix multiplication is not commutative. Let \(T\) represent a matrix holding the final positions of the image (after the transformation).

     \[T = RP\]

     After you calculate \(T\), you can plot the points in \(T\) with the following code (in SageMath):

     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     Which is the graph of the points in \(T\)? (Please excuse minor stylistic differences due to my graphs being made in R, not SageMath.)

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     P = matrix([[-2,-1,0,4,-4,-6,-3,2,6,7,-1,-1],[5,-2,-7,7,6,2,-7,-4,-1,-3,6,-6]])
     theta = 1.42
     R = matrix([[cos(theta),-sin(theta)],
                 [sin(theta),cos(theta)]])
     T = R*P
     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     
     
119. 
     

     Question

     prevnext

     Matrix of original points (preimage)

     A list of 2D-cartesian points is stored as the preimage matrix. Each column represents one point.

     \[P = \begin{bmatrix} 0 & -3 & -4 & -4 & 6 & -4 & 3 & 4 & -3 & 6\\ 3 & -1 & -2 & -4 & -4 & -5 & -2 & -7 & -5 & -7 \end{bmatrix}\]

     P = matrix([[0,-3,-4,-4,6,-4,3,4,-3,6],[3,-1,-2,-4,-4,-5,-2,-7,-5,-7]])

     Rotation matrix

     We want to rotate the points by \(\theta=0.88\) radians counterclockwise. We will do this by matrix multiplication with the 2D rotation matrix.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\]

     Image

     We can apply the rotation to the points through matrix multiplication. Remember, matrix multiplication is not commutative. Let \(T\) represent a matrix holding the final positions of the image (after the transformation).

     \[T = RP\]

     After you calculate \(T\), you can plot the points in \(T\) with the following code (in SageMath):

     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     Which is the graph of the points in \(T\)? (Please excuse minor stylistic differences due to my graphs being made in R, not SageMath.)

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     P = matrix([[0,-3,-4,-4,6,-4,3,4,-3,6],[3,-1,-2,-4,-4,-5,-2,-7,-5,-7]])
     theta = 0.88
     R = matrix([[cos(theta),-sin(theta)],
                 [sin(theta),cos(theta)]])
     T = R*P
     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     
     
120. 
     

     Question

     prevnext

     Matrix of original points (preimage)

     A list of 2D-cartesian points is stored as the preimage matrix. Each column represents one point.

     \[P = \begin{bmatrix} 7 & -5 & -4 & 7 & 4 & -3 & 1 & -6 & -2 & 4 & 3 & -5\\ 6 & -2 & 5 & 1 & -5 & -5 & -6 & 7 & -2 & 0 & 7 & -3 \end{bmatrix}\]

     P = matrix([[7,-5,-4,7,4,-3,1,-6,-2,4,3,-5],[6,-2,5,1,-5,-5,-6,7,-2,0,7,-3]])

     Rotation matrix

     We want to rotate the points by \(\theta=4.12\) radians counterclockwise. We will do this by matrix multiplication with the 2D rotation matrix.

     \[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\]

     Image

     We can apply the rotation to the points through matrix multiplication. Remember, matrix multiplication is not commutative. Let \(T\) represent a matrix holding the final positions of the image (after the transformation).

     \[T = RP\]

     After you calculate \(T\), you can plot the points in \(T\) with the following code (in SageMath):

     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     Which is the graph of the points in \(T\)? (Please excuse minor stylistic differences due to my graphs being made in R, not SageMath.)

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     P = matrix([[7,-5,-4,7,4,-3,1,-6,-2,4,3,-5],[6,-2,5,1,-5,-5,-6,7,-2,0,7,-3]])
     theta = 4.12
     R = matrix([[cos(theta),-sin(theta)],
                 [sin(theta),cos(theta)]])
     T = R*P
     point(T.transpose(),xmin=-10,xmax=10,ymin=-10,ymax=10,size=50)

     
     
121. 
     

     Question

     prevnext

     Some basic 2D-transformation matrices are shown:

     • Horizontal stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}k & 0 \\ 0 & 1\end{bmatrix}\]
     • Vertical stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}1 & 0 \\ 0 & k\end{bmatrix}\]
     • Rotation by angle \(\theta\) radians. \[\begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}\]
     • Horizontal shearing by factor \(k\). \[\begin{bmatrix}1 & k \\ 0 & 1\end{bmatrix}\]
     • Vertical shearing by factor \(k\). \[\begin{bmatrix}1 & 0 \\ k & 1\end{bmatrix}\]

     When multiple transformations are applied, order matters. If \(A\) was the matrix of the first transformation, \(B\) was the matrix of the second transformation, \(C\) was the matrix of the third transformation, and \(X\) was a matrix of column vectors, then the result, \(X'\), would be found with the following matrix multiplication: \[X' = CBAX\] Notice, the order seems backwards; this is standard.

     Problem-specific information

     Begin with the points of a unit square.

     \[X = \begin{bmatrix}0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}\]

     X = matrix([[0,1,1,0],[0,0,1,1]])
     polygon(X.transpose(),xmin=-5,xmax=5,ymin=-5,ymax=5,aspect_ratio=1,color="red")

     Apply the following transformations in the indicated order:

     1. A rotation by angle 4.24.
     2. A vertical shear by factor 2.2.
     3. A horizontal shear by factor 1.5.

     Which graph is generated when polygon \(X\) is plotted in red and polygon \(X'\) is plotted in blue?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is C.

     SageMath

     mygraph = Graphics()
     X = matrix([[0,1,1,0],[0,0,1,1]])
     mygraph += polygon(X.transpose(),color="red",alpha=0.5)
     A = matrix([[-0.455,0.89],
                 [-0.89,-0.455]])
     B = matrix([[1,0],
                 [2.2,1]])
     C = matrix([[1,1.5],
                 [0,1]])
     XP = C*B*A*X
     mygraph += polygon(XP.transpose(),color="blue",alpha=0.5)
     show(mygraph,xmin=-10,xmax=10,ymin=-10,ymax=10,aspect_ratio=1)

     
     
122. 
     

     Question

     prevnext

     Some basic 2D-transformation matrices are shown:

     • Horizontal stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}k & 0 \\ 0 & 1\end{bmatrix}\]
     • Vertical stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}1 & 0 \\ 0 & k\end{bmatrix}\]
     • Rotation by angle \(\theta\) radians. \[\begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}\]
     • Horizontal shearing by factor \(k\). \[\begin{bmatrix}1 & k \\ 0 & 1\end{bmatrix}\]
     • Vertical shearing by factor \(k\). \[\begin{bmatrix}1 & 0 \\ k & 1\end{bmatrix}\]

     When multiple transformations are applied, order matters. If \(A\) was the matrix of the first transformation, \(B\) was the matrix of the second transformation, \(C\) was the matrix of the third transformation, and \(X\) was a matrix of column vectors, then the result, \(X'\), would be found with the following matrix multiplication: \[X' = CBAX\] Notice, the order seems backwards; this is standard.

     Problem-specific information

     Begin with the points of a unit square.

     \[X = \begin{bmatrix}0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}\]

     X = matrix([[0,1,1,0],[0,0,1,1]])
     polygon(X.transpose(),xmin=-5,xmax=5,ymin=-5,ymax=5,aspect_ratio=1,color="red")

     Apply the following transformations in the indicated order:

     1. A horizontal stretch by factor 1.9.
     2. A vertical shear by factor 2.3.
     3. A rotation by angle 4.18.

     Which graph is generated when polygon \(X\) is plotted in red and polygon \(X'\) is plotted in blue?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is A.

     SageMath

     mygraph = Graphics()
     X = matrix([[0,1,1,0],[0,0,1,1]])
     mygraph += polygon(X.transpose(),color="red",alpha=0.5)
     A = matrix([[1.9,0],
                 [0,1]])
     B = matrix([[1,0],
                 [2.3,1]])
     C = matrix([[-0.508,0.862],
                 [-0.862,-0.508]])
     XP = C*B*A*X
     mygraph += polygon(XP.transpose(),color="blue",alpha=0.5)
     show(mygraph,xmin=-10,xmax=10,ymin=-10,ymax=10,aspect_ratio=1)

     
     
123. 
     

     Question

     prevnext

     Some basic 2D-transformation matrices are shown:

     • Horizontal stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}k & 0 \\ 0 & 1\end{bmatrix}\]
     • Vertical stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}1 & 0 \\ 0 & k\end{bmatrix}\]
     • Rotation by angle \(\theta\) radians. \[\begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}\]
     • Horizontal shearing by factor \(k\). \[\begin{bmatrix}1 & k \\ 0 & 1\end{bmatrix}\]
     • Vertical shearing by factor \(k\). \[\begin{bmatrix}1 & 0 \\ k & 1\end{bmatrix}\]

     When multiple transformations are applied, order matters. If \(A\) was the matrix of the first transformation, \(B\) was the matrix of the second transformation, \(C\) was the matrix of the third transformation, and \(X\) was a matrix of column vectors, then the result, \(X'\), would be found with the following matrix multiplication: \[X' = CBAX\] Notice, the order seems backwards; this is standard.

     Problem-specific information

     Begin with the points of a unit square.

     \[X = \begin{bmatrix}0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}\]

     X = matrix([[0,1,1,0],[0,0,1,1]])
     polygon(X.transpose(),xmin=-5,xmax=5,ymin=-5,ymax=5,aspect_ratio=1,color="red")

     Apply the following transformations in the indicated order:

     1. A horizontal shear by factor 1.6.
     2. A vertical shear by factor 2.2.
     3. A rotation by angle 4.89.

     Which graph is generated when polygon \(X\) is plotted in red and polygon \(X'\) is plotted in blue?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is B.

     SageMath

     mygraph = Graphics()
     X = matrix([[0,1,1,0],[0,0,1,1]])
     mygraph += polygon(X.transpose(),color="red",alpha=0.5)
     A = matrix([[1,1.6],
                 [0,1]])
     B = matrix([[1,0],
                 [2.2,1]])
     C = matrix([[0.177,0.984],
                 [-0.984,0.177]])
     XP = C*B*A*X
     mygraph += polygon(XP.transpose(),color="blue",alpha=0.5)
     show(mygraph,xmin=-10,xmax=10,ymin=-10,ymax=10,aspect_ratio=1)

     
     
124. 
     

     Question

     prevnext

     Some basic 2D-transformation matrices are shown:

     • Horizontal stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}k & 0 \\ 0 & 1\end{bmatrix}\]
     • Vertical stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}1 & 0 \\ 0 & k\end{bmatrix}\]
     • Rotation by angle \(\theta\) radians. \[\begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}\]
     • Horizontal shearing by factor \(k\). \[\begin{bmatrix}1 & k \\ 0 & 1\end{bmatrix}\]
     • Vertical shearing by factor \(k\). \[\begin{bmatrix}1 & 0 \\ k & 1\end{bmatrix}\]

     When multiple transformations are applied, order matters. If \(A\) was the matrix of the first transformation, \(B\) was the matrix of the second transformation, \(C\) was the matrix of the third transformation, and \(X\) was a matrix of column vectors, then the result, \(X'\), would be found with the following matrix multiplication: \[X' = CBAX\] Notice, the order seems backwards; this is standard.

     Problem-specific information

     Begin with the points of a unit square.

     \[X = \begin{bmatrix}0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}\]

     X = matrix([[0,1,1,0],[0,0,1,1]])
     polygon(X.transpose(),xmin=-5,xmax=5,ymin=-5,ymax=5,aspect_ratio=1,color="red")

     Apply the following transformations in the indicated order:

     1. A horizontal stretch by factor 2.3.
     2. A rotation by angle 0.4.
     3. A vertical stretch by factor 1.7.

     Which graph is generated when polygon \(X\) is plotted in red and polygon \(X'\) is plotted in blue?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is A.

     SageMath

     mygraph = Graphics()
     X = matrix([[0,1,1,0],[0,0,1,1]])
     mygraph += polygon(X.transpose(),color="red",alpha=0.5)
     A = matrix([[2.3,0],
                 [0,1]])
     B = matrix([[0.921,-0.389],
                 [0.389,0.921]])
     C = matrix([[1,0],
                 [0,1.7]])
     XP = C*B*A*X
     mygraph += polygon(XP.transpose(),color="blue",alpha=0.5)
     show(mygraph,xmin=-10,xmax=10,ymin=-10,ymax=10,aspect_ratio=1)

     
     
125. 
     

     Question

     prevnext

     Some basic 2D-transformation matrices are shown:

     • Horizontal stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}k & 0 \\ 0 & 1\end{bmatrix}\]
     • Vertical stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}1 & 0 \\ 0 & k\end{bmatrix}\]
     • Rotation by angle \(\theta\) radians. \[\begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}\]
     • Horizontal shearing by factor \(k\). \[\begin{bmatrix}1 & k \\ 0 & 1\end{bmatrix}\]
     • Vertical shearing by factor \(k\). \[\begin{bmatrix}1 & 0 \\ k & 1\end{bmatrix}\]

     When multiple transformations are applied, order matters. If \(A\) was the matrix of the first transformation, \(B\) was the matrix of the second transformation, \(C\) was the matrix of the third transformation, and \(X\) was a matrix of column vectors, then the result, \(X'\), would be found with the following matrix multiplication: \[X' = CBAX\] Notice, the order seems backwards; this is standard.

     Problem-specific information

     Begin with the points of a unit square.

     \[X = \begin{bmatrix}0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}\]

     X = matrix([[0,1,1,0],[0,0,1,1]])
     polygon(X.transpose(),xmin=-5,xmax=5,ymin=-5,ymax=5,aspect_ratio=1,color="red")

     Apply the following transformations in the indicated order:

     1. A vertical shear by factor 2.2.
     2. A rotation by angle 0.53.
     3. A horizontal shear by factor 1.3.

     Which graph is generated when polygon \(X\) is plotted in red and polygon \(X'\) is plotted in blue?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is A.

     SageMath

     mygraph = Graphics()
     X = matrix([[0,1,1,0],[0,0,1,1]])
     mygraph += polygon(X.transpose(),color="red",alpha=0.5)
     A = matrix([[1,0],
                 [2.2,1]])
     B = matrix([[0.863,-0.506],
                 [0.506,0.863]])
     C = matrix([[1,1.3],
                 [0,1]])
     XP = C*B*A*X
     mygraph += polygon(XP.transpose(),color="blue",alpha=0.5)
     show(mygraph,xmin=-10,xmax=10,ymin=-10,ymax=10,aspect_ratio=1)

     
     
126. 
     

     Question

     prevnext

     Some basic 2D-transformation matrices are shown:

     • Horizontal stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}k & 0 \\ 0 & 1\end{bmatrix}\]
     • Vertical stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}1 & 0 \\ 0 & k\end{bmatrix}\]
     • Rotation by angle \(\theta\) radians. \[\begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}\]
     • Horizontal shearing by factor \(k\). \[\begin{bmatrix}1 & k \\ 0 & 1\end{bmatrix}\]
     • Vertical shearing by factor \(k\). \[\begin{bmatrix}1 & 0 \\ k & 1\end{bmatrix}\]

     When multiple transformations are applied, order matters. If \(A\) was the matrix of the first transformation, \(B\) was the matrix of the second transformation, \(C\) was the matrix of the third transformation, and \(X\) was a matrix of column vectors, then the result, \(X'\), would be found with the following matrix multiplication: \[X' = CBAX\] Notice, the order seems backwards; this is standard.

     Problem-specific information

     Begin with the points of a unit square.

     \[X = \begin{bmatrix}0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}\]

     X = matrix([[0,1,1,0],[0,0,1,1]])
     polygon(X.transpose(),xmin=-5,xmax=5,ymin=-5,ymax=5,aspect_ratio=1,color="red")

     Apply the following transformations in the indicated order:

     1. A vertical shear by factor 1.5.
     2. A vertical stretch by factor 1.3.
     3. A rotation by angle 2.79.

     Which graph is generated when polygon \(X\) is plotted in red and polygon \(X'\) is plotted in blue?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is A.

     SageMath

     mygraph = Graphics()
     X = matrix([[0,1,1,0],[0,0,1,1]])
     mygraph += polygon(X.transpose(),color="red",alpha=0.5)
     A = matrix([[1,0],
                 [1.5,1]])
     B = matrix([[1,0],
                 [0,1.3]])
     C = matrix([[-0.939,-0.344],
                 [0.344,-0.939]])
     XP = C*B*A*X
     mygraph += polygon(XP.transpose(),color="blue",alpha=0.5)
     show(mygraph,xmin=-10,xmax=10,ymin=-10,ymax=10,aspect_ratio=1)

     
     
127. 
     

     Question

     prevnext

     Some basic 2D-transformation matrices are shown:

     • Horizontal stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}k & 0 \\ 0 & 1\end{bmatrix}\]
     • Vertical stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}1 & 0 \\ 0 & k\end{bmatrix}\]
     • Rotation by angle \(\theta\) radians. \[\begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}\]
     • Horizontal shearing by factor \(k\). \[\begin{bmatrix}1 & k \\ 0 & 1\end{bmatrix}\]
     • Vertical shearing by factor \(k\). \[\begin{bmatrix}1 & 0 \\ k & 1\end{bmatrix}\]

     When multiple transformations are applied, order matters. If \(A\) was the matrix of the first transformation, \(B\) was the matrix of the second transformation, \(C\) was the matrix of the third transformation, and \(X\) was a matrix of column vectors, then the result, \(X'\), would be found with the following matrix multiplication: \[X' = CBAX\] Notice, the order seems backwards; this is standard.

     Problem-specific information

     Begin with the points of a unit square.

     \[X = \begin{bmatrix}0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}\]

     X = matrix([[0,1,1,0],[0,0,1,1]])
     polygon(X.transpose(),xmin=-5,xmax=5,ymin=-5,ymax=5,aspect_ratio=1,color="red")

     Apply the following transformations in the indicated order:

     1. A rotation by angle 3.41.
     2. A horizontal shear by factor 2.2.
     3. A vertical shear by factor 1.7.

     Which graph is generated when polygon \(X\) is plotted in red and polygon \(X'\) is plotted in blue?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is D.

     SageMath

     mygraph = Graphics()
     X = matrix([[0,1,1,0],[0,0,1,1]])
     mygraph += polygon(X.transpose(),color="red",alpha=0.5)
     A = matrix([[-0.964,0.265],
                 [-0.265,-0.964]])
     B = matrix([[1,2.2],
                 [0,1]])
     C = matrix([[1,0],
                 [1.7,1]])
     XP = C*B*A*X
     mygraph += polygon(XP.transpose(),color="blue",alpha=0.5)
     show(mygraph,xmin=-10,xmax=10,ymin=-10,ymax=10,aspect_ratio=1)

     
     
128. 
     

     Question

     prevnext

     Some basic 2D-transformation matrices are shown:

     • Horizontal stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}k & 0 \\ 0 & 1\end{bmatrix}\]
     • Vertical stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}1 & 0 \\ 0 & k\end{bmatrix}\]
     • Rotation by angle \(\theta\) radians. \[\begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}\]
     • Horizontal shearing by factor \(k\). \[\begin{bmatrix}1 & k \\ 0 & 1\end{bmatrix}\]
     • Vertical shearing by factor \(k\). \[\begin{bmatrix}1 & 0 \\ k & 1\end{bmatrix}\]

     When multiple transformations are applied, order matters. If \(A\) was the matrix of the first transformation, \(B\) was the matrix of the second transformation, \(C\) was the matrix of the third transformation, and \(X\) was a matrix of column vectors, then the result, \(X'\), would be found with the following matrix multiplication: \[X' = CBAX\] Notice, the order seems backwards; this is standard.

     Problem-specific information

     Begin with the points of a unit square.

     \[X = \begin{bmatrix}0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}\]

     X = matrix([[0,1,1,0],[0,0,1,1]])
     polygon(X.transpose(),xmin=-5,xmax=5,ymin=-5,ymax=5,aspect_ratio=1,color="red")

     Apply the following transformations in the indicated order:

     1. A rotation by angle 2.46.
     2. A vertical shear by factor 1.8.
     3. A horizontal shear by factor 2.3.

     Which graph is generated when polygon \(X\) is plotted in red and polygon \(X'\) is plotted in blue?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is C.

     SageMath

     mygraph = Graphics()
     X = matrix([[0,1,1,0],[0,0,1,1]])
     mygraph += polygon(X.transpose(),color="red",alpha=0.5)
     A = matrix([[-0.777,-0.63],
                 [0.63,-0.777]])
     B = matrix([[1,0],
                 [1.8,1]])
     C = matrix([[1,2.3],
                 [0,1]])
     XP = C*B*A*X
     mygraph += polygon(XP.transpose(),color="blue",alpha=0.5)
     show(mygraph,xmin=-10,xmax=10,ymin=-10,ymax=10,aspect_ratio=1)

     
     
129. 
     

     Question

     prevnext

     Some basic 2D-transformation matrices are shown:

     • Horizontal stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}k & 0 \\ 0 & 1\end{bmatrix}\]
     • Vertical stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}1 & 0 \\ 0 & k\end{bmatrix}\]
     • Rotation by angle \(\theta\) radians. \[\begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}\]
     • Horizontal shearing by factor \(k\). \[\begin{bmatrix}1 & k \\ 0 & 1\end{bmatrix}\]
     • Vertical shearing by factor \(k\). \[\begin{bmatrix}1 & 0 \\ k & 1\end{bmatrix}\]

     When multiple transformations are applied, order matters. If \(A\) was the matrix of the first transformation, \(B\) was the matrix of the second transformation, \(C\) was the matrix of the third transformation, and \(X\) was a matrix of column vectors, then the result, \(X'\), would be found with the following matrix multiplication: \[X' = CBAX\] Notice, the order seems backwards; this is standard.

     Problem-specific information

     Begin with the points of a unit square.

     \[X = \begin{bmatrix}0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}\]

     X = matrix([[0,1,1,0],[0,0,1,1]])
     polygon(X.transpose(),xmin=-5,xmax=5,ymin=-5,ymax=5,aspect_ratio=1,color="red")

     Apply the following transformations in the indicated order:

     1. A rotation by angle 2.03.
     2. A vertical stretch by factor 1.6.
     3. A horizontal stretch by factor 1.4.

     Which graph is generated when polygon \(X\) is plotted in red and polygon \(X'\) is plotted in blue?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is B.

     SageMath

     mygraph = Graphics()
     X = matrix([[0,1,1,0],[0,0,1,1]])
     mygraph += polygon(X.transpose(),color="red",alpha=0.5)
     A = matrix([[-0.443,-0.896],
                 [0.896,-0.443]])
     B = matrix([[1,0],
                 [0,1.6]])
     C = matrix([[1.4,0],
                 [0,1]])
     XP = C*B*A*X
     mygraph += polygon(XP.transpose(),color="blue",alpha=0.5)
     show(mygraph,xmin=-10,xmax=10,ymin=-10,ymax=10,aspect_ratio=1)

     
     
130. 
     

     Question

     prevnext

     Some basic 2D-transformation matrices are shown:

     • Horizontal stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}k & 0 \\ 0 & 1\end{bmatrix}\]
     • Vertical stretch by factor \(k\) (reflection if \(k=-1\)). \[\begin{bmatrix}1 & 0 \\ 0 & k\end{bmatrix}\]
     • Rotation by angle \(\theta\) radians. \[\begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}\]
     • Horizontal shearing by factor \(k\). \[\begin{bmatrix}1 & k \\ 0 & 1\end{bmatrix}\]
     • Vertical shearing by factor \(k\). \[\begin{bmatrix}1 & 0 \\ k & 1\end{bmatrix}\]

     When multiple transformations are applied, order matters. If \(A\) was the matrix of the first transformation, \(B\) was the matrix of the second transformation, \(C\) was the matrix of the third transformation, and \(X\) was a matrix of column vectors, then the result, \(X'\), would be found with the following matrix multiplication: \[X' = CBAX\] Notice, the order seems backwards; this is standard.

     Problem-specific information

     Begin with the points of a unit square.

     \[X = \begin{bmatrix}0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}\]

     X = matrix([[0,1,1,0],[0,0,1,1]])
     polygon(X.transpose(),xmin=-5,xmax=5,ymin=-5,ymax=5,aspect_ratio=1,color="red")

     Apply the following transformations in the indicated order:

     1. A vertical shear by factor 2.3.
     2. A vertical stretch by factor 1.9.
     3. A rotation by angle 3.03.

     Which graph is generated when polygon \(X\) is plotted in red and polygon \(X'\) is plotted in blue?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is A.

     SageMath

     mygraph = Graphics()
     X = matrix([[0,1,1,0],[0,0,1,1]])
     mygraph += polygon(X.transpose(),color="red",alpha=0.5)
     A = matrix([[1,0],
                 [2.3,1]])
     B = matrix([[1,0],
                 [0,1.9]])
     C = matrix([[-0.994,-0.111],
                 [0.111,-0.994]])
     XP = C*B*A*X
     mygraph += polygon(XP.transpose(),color="blue",alpha=0.5)
     show(mygraph,xmin=-10,xmax=10,ymin=-10,ymax=10,aspect_ratio=1)

     
     
131. 
     

     Question

     prevnext

     One of the simplest 3D \(\to\) 2D projections (to draw by hand) is an oblique projection. Each type of oblique projection can be parameterized with an angle (\(\theta\)) and a ratio (\(r<1\)). Below is an oblique-projection matrix.

     \[P = \begin{bmatrix}1 & r\cos(\theta_P) & 0 \\ 0 & r\sin(\theta_P) & 1 \end{bmatrix}\]

     We can exemplify an oblique projection with \(r=0.5\) and \(\theta_P=\frac{pi}{6}\) on a unit cube. The cube has lengths 1 unit long.

     (SageMath)

     X1 = matrix([[0,0,0,0,0,1,1,0,1,1,0,1,1,0,1,1],[0,0,1,1,0,0,1,1,1,1,1,1,0,0,0,0],[0,1,1,0,0,0,0,0,0,1,1,1,1,1,1,0]])
     r = 0.5
     tp = pi/6
     P = matrix([[1,r*cos(tp),0],[0,r*sin(tp),1]])
     flat_X1 = P*X1
     line(flat_X1.transpose(),aspect_ratio=1)

     
     

     In 3D, there are relatively simple rotation matrices for spinning around the \(x\), \(y\), or \(z\) axis.

     \[R_x = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos(\theta_x) & -\sin(\theta_x) \\ 0 &\sin(\theta_x) & \cos(\theta_x) \end{bmatrix}\]

     \[R_y = \begin{bmatrix}\cos(\theta_y) & 0 & \sin(\theta_y) \\ 0 & 1 & 0 \\ -\sin(\theta_y) & 0 & \cos(\theta_y) \end{bmatrix}\]

     \[R_z = \begin{bmatrix}\cos(\theta_z) & -\sin(\theta_z) & 0 \\ \sin(\theta_z) & \cos(\theta_z) & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     Information specific to this question

     Fill in the missing code to generate various rotations of a square pyramid.

     • Red: X1 is the original
     • Orange: X2 is after a rotation of 0.87 radians around the x-axis
     • Green: X3 has another rotation of 3.63 radians around the y-axis
     • Blue: X4 has another rotation of 2.45 radians around the z-axis

     In order to flatten the 3D points onto a 2D graph, each pyramid will undergo an oblique projection with \(r=0.5\) and \(\theta_P=0.33\).

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = ...
     X1_flat = ...
     graph01 += line(X1_flat.transpose(),color="red")
     Rx = ...
     X2 = ...
     X2_flat = ...
     graph01 += line(X2_flat.transpose(),color="orange")
     Ry = ...
     X3 = ...
     X3_flat = ...
     graph01 += ...
     Rz = ...
     X4 = ...
     X4_flat = ...
     graph01 += ...
     show(graph01,aspect_ratio=1,axes=False)

     Which graph matches?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is C.

     SageMath

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = matrix([[1,0.5*cos(0.33),0],[0,0.5*sin(0.33),1]])
     X1_flat = P*X1
     graph01 += line(X1_flat.transpose(),color="red")
     tx = 0.87
     Rx = matrix([[1,0,0],[0,cos(tx),-sin(tx)],[0,sin(tx),cos(tx)]])
     X2 = Rx*X1
     X2_flat = P*X2
     graph01 += line(X2_flat.transpose(),color="orange")
     ty = 3.63
     Ry = matrix([[cos(ty),0,sin(ty)],[0,1,0],[-sin(ty),0,cos(ty)]])
     X3 = Ry*X2
     X3_flat = P*X3
     graph01 += line(X3_flat.transpose(),color="green")
     Rz = matrix([[cos(2.45),-sin(2.45),0],[sin(2.45),cos(2.45),0],[0,0,1]])
     X4 = Rz*X3
     X4_flat = P*X4
     graph01 += line(X4_flat.transpose(),color="blue")
     show(graph01,aspect_ratio=1,axes=False)

     
     
132. 
     

     Question

     prevnext

     One of the simplest 3D \(\to\) 2D projections (to draw by hand) is an oblique projection. Each type of oblique projection can be parameterized with an angle (\(\theta\)) and a ratio (\(r<1\)). Below is an oblique-projection matrix.

     \[P = \begin{bmatrix}1 & r\cos(\theta_P) & 0 \\ 0 & r\sin(\theta_P) & 1 \end{bmatrix}\]

     We can exemplify an oblique projection with \(r=0.5\) and \(\theta_P=\frac{pi}{6}\) on a unit cube. The cube has lengths 1 unit long.

     (SageMath)

     X1 = matrix([[0,0,0,0,0,1,1,0,1,1,0,1,1,0,1,1],[0,0,1,1,0,0,1,1,1,1,1,1,0,0,0,0],[0,1,1,0,0,0,0,0,0,1,1,1,1,1,1,0]])
     r = 0.5
     tp = pi/6
     P = matrix([[1,r*cos(tp),0],[0,r*sin(tp),1]])
     flat_X1 = P*X1
     line(flat_X1.transpose(),aspect_ratio=1)

     
     

     In 3D, there are relatively simple rotation matrices for spinning around the \(x\), \(y\), or \(z\) axis.

     \[R_x = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos(\theta_x) & -\sin(\theta_x) \\ 0 &\sin(\theta_x) & \cos(\theta_x) \end{bmatrix}\]

     \[R_y = \begin{bmatrix}\cos(\theta_y) & 0 & \sin(\theta_y) \\ 0 & 1 & 0 \\ -\sin(\theta_y) & 0 & \cos(\theta_y) \end{bmatrix}\]

     \[R_z = \begin{bmatrix}\cos(\theta_z) & -\sin(\theta_z) & 0 \\ \sin(\theta_z) & \cos(\theta_z) & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     Information specific to this question

     Fill in the missing code to generate various rotations of a square pyramid.

     • Red: X1 is the original
     • Orange: X2 is after a rotation of 5.22 radians around the x-axis
     • Green: X3 has another rotation of 4.59 radians around the y-axis
     • Blue: X4 has another rotation of 1.08 radians around the z-axis

     In order to flatten the 3D points onto a 2D graph, each pyramid will undergo an oblique projection with \(r=0.5\) and \(\theta_P=0.71\).

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = ...
     X1_flat = ...
     graph01 += line(X1_flat.transpose(),color="red")
     Rx = ...
     X2 = ...
     X2_flat = ...
     graph01 += line(X2_flat.transpose(),color="orange")
     Ry = ...
     X3 = ...
     X3_flat = ...
     graph01 += ...
     Rz = ...
     X4 = ...
     X4_flat = ...
     graph01 += ...
     show(graph01,aspect_ratio=1,axes=False)

     Which graph matches?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is D.

     SageMath

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = matrix([[1,0.5*cos(0.71),0],[0,0.5*sin(0.71),1]])
     X1_flat = P*X1
     graph01 += line(X1_flat.transpose(),color="red")
     tx = 5.22
     Rx = matrix([[1,0,0],[0,cos(tx),-sin(tx)],[0,sin(tx),cos(tx)]])
     X2 = Rx*X1
     X2_flat = P*X2
     graph01 += line(X2_flat.transpose(),color="orange")
     ty = 4.59
     Ry = matrix([[cos(ty),0,sin(ty)],[0,1,0],[-sin(ty),0,cos(ty)]])
     X3 = Ry*X2
     X3_flat = P*X3
     graph01 += line(X3_flat.transpose(),color="green")
     Rz = matrix([[cos(1.08),-sin(1.08),0],[sin(1.08),cos(1.08),0],[0,0,1]])
     X4 = Rz*X3
     X4_flat = P*X4
     graph01 += line(X4_flat.transpose(),color="blue")
     show(graph01,aspect_ratio=1,axes=False)

     
     
133. 
     

     Question

     prevnext

     One of the simplest 3D \(\to\) 2D projections (to draw by hand) is an oblique projection. Each type of oblique projection can be parameterized with an angle (\(\theta\)) and a ratio (\(r<1\)). Below is an oblique-projection matrix.

     \[P = \begin{bmatrix}1 & r\cos(\theta_P) & 0 \\ 0 & r\sin(\theta_P) & 1 \end{bmatrix}\]

     We can exemplify an oblique projection with \(r=0.5\) and \(\theta_P=\frac{pi}{6}\) on a unit cube. The cube has lengths 1 unit long.

     (SageMath)

     X1 = matrix([[0,0,0,0,0,1,1,0,1,1,0,1,1,0,1,1],[0,0,1,1,0,0,1,1,1,1,1,1,0,0,0,0],[0,1,1,0,0,0,0,0,0,1,1,1,1,1,1,0]])
     r = 0.5
     tp = pi/6
     P = matrix([[1,r*cos(tp),0],[0,r*sin(tp),1]])
     flat_X1 = P*X1
     line(flat_X1.transpose(),aspect_ratio=1)

     
     

     In 3D, there are relatively simple rotation matrices for spinning around the \(x\), \(y\), or \(z\) axis.

     \[R_x = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos(\theta_x) & -\sin(\theta_x) \\ 0 &\sin(\theta_x) & \cos(\theta_x) \end{bmatrix}\]

     \[R_y = \begin{bmatrix}\cos(\theta_y) & 0 & \sin(\theta_y) \\ 0 & 1 & 0 \\ -\sin(\theta_y) & 0 & \cos(\theta_y) \end{bmatrix}\]

     \[R_z = \begin{bmatrix}\cos(\theta_z) & -\sin(\theta_z) & 0 \\ \sin(\theta_z) & \cos(\theta_z) & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     Information specific to this question

     Fill in the missing code to generate various rotations of a square pyramid.

     • Red: X1 is the original
     • Orange: X2 is after a rotation of 3.61 radians around the x-axis
     • Green: X3 has another rotation of 0.44 radians around the y-axis
     • Blue: X4 has another rotation of 0.48 radians around the z-axis

     In order to flatten the 3D points onto a 2D graph, each pyramid will undergo an oblique projection with \(r=0.5\) and \(\theta_P=0.39\).

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = ...
     X1_flat = ...
     graph01 += line(X1_flat.transpose(),color="red")
     Rx = ...
     X2 = ...
     X2_flat = ...
     graph01 += line(X2_flat.transpose(),color="orange")
     Ry = ...
     X3 = ...
     X3_flat = ...
     graph01 += ...
     Rz = ...
     X4 = ...
     X4_flat = ...
     graph01 += ...
     show(graph01,aspect_ratio=1,axes=False)

     Which graph matches?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is B.

     SageMath

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = matrix([[1,0.5*cos(0.39),0],[0,0.5*sin(0.39),1]])
     X1_flat = P*X1
     graph01 += line(X1_flat.transpose(),color="red")
     tx = 3.61
     Rx = matrix([[1,0,0],[0,cos(tx),-sin(tx)],[0,sin(tx),cos(tx)]])
     X2 = Rx*X1
     X2_flat = P*X2
     graph01 += line(X2_flat.transpose(),color="orange")
     ty = 0.44
     Ry = matrix([[cos(ty),0,sin(ty)],[0,1,0],[-sin(ty),0,cos(ty)]])
     X3 = Ry*X2
     X3_flat = P*X3
     graph01 += line(X3_flat.transpose(),color="green")
     Rz = matrix([[cos(0.48),-sin(0.48),0],[sin(0.48),cos(0.48),0],[0,0,1]])
     X4 = Rz*X3
     X4_flat = P*X4
     graph01 += line(X4_flat.transpose(),color="blue")
     show(graph01,aspect_ratio=1,axes=False)

     
     
134. 
     

     Question

     prevnext

     One of the simplest 3D \(\to\) 2D projections (to draw by hand) is an oblique projection. Each type of oblique projection can be parameterized with an angle (\(\theta\)) and a ratio (\(r<1\)). Below is an oblique-projection matrix.

     \[P = \begin{bmatrix}1 & r\cos(\theta_P) & 0 \\ 0 & r\sin(\theta_P) & 1 \end{bmatrix}\]

     We can exemplify an oblique projection with \(r=0.5\) and \(\theta_P=\frac{pi}{6}\) on a unit cube. The cube has lengths 1 unit long.

     (SageMath)

     X1 = matrix([[0,0,0,0,0,1,1,0,1,1,0,1,1,0,1,1],[0,0,1,1,0,0,1,1,1,1,1,1,0,0,0,0],[0,1,1,0,0,0,0,0,0,1,1,1,1,1,1,0]])
     r = 0.5
     tp = pi/6
     P = matrix([[1,r*cos(tp),0],[0,r*sin(tp),1]])
     flat_X1 = P*X1
     line(flat_X1.transpose(),aspect_ratio=1)

     
     

     In 3D, there are relatively simple rotation matrices for spinning around the \(x\), \(y\), or \(z\) axis.

     \[R_x = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos(\theta_x) & -\sin(\theta_x) \\ 0 &\sin(\theta_x) & \cos(\theta_x) \end{bmatrix}\]

     \[R_y = \begin{bmatrix}\cos(\theta_y) & 0 & \sin(\theta_y) \\ 0 & 1 & 0 \\ -\sin(\theta_y) & 0 & \cos(\theta_y) \end{bmatrix}\]

     \[R_z = \begin{bmatrix}\cos(\theta_z) & -\sin(\theta_z) & 0 \\ \sin(\theta_z) & \cos(\theta_z) & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     Information specific to this question

     Fill in the missing code to generate various rotations of a square pyramid.

     • Red: X1 is the original
     • Orange: X2 is after a rotation of 2.17 radians around the x-axis
     • Green: X3 has another rotation of 3 radians around the y-axis
     • Blue: X4 has another rotation of 0.5 radians around the z-axis

     In order to flatten the 3D points onto a 2D graph, each pyramid will undergo an oblique projection with \(r=0.5\) and \(\theta_P=0.65\).

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = ...
     X1_flat = ...
     graph01 += line(X1_flat.transpose(),color="red")
     Rx = ...
     X2 = ...
     X2_flat = ...
     graph01 += line(X2_flat.transpose(),color="orange")
     Ry = ...
     X3 = ...
     X3_flat = ...
     graph01 += ...
     Rz = ...
     X4 = ...
     X4_flat = ...
     graph01 += ...
     show(graph01,aspect_ratio=1,axes=False)

     Which graph matches?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is C.

     SageMath

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = matrix([[1,0.5*cos(0.65),0],[0,0.5*sin(0.65),1]])
     X1_flat = P*X1
     graph01 += line(X1_flat.transpose(),color="red")
     tx = 2.17
     Rx = matrix([[1,0,0],[0,cos(tx),-sin(tx)],[0,sin(tx),cos(tx)]])
     X2 = Rx*X1
     X2_flat = P*X2
     graph01 += line(X2_flat.transpose(),color="orange")
     ty = 3
     Ry = matrix([[cos(ty),0,sin(ty)],[0,1,0],[-sin(ty),0,cos(ty)]])
     X3 = Ry*X2
     X3_flat = P*X3
     graph01 += line(X3_flat.transpose(),color="green")
     Rz = matrix([[cos(0.5),-sin(0.5),0],[sin(0.5),cos(0.5),0],[0,0,1]])
     X4 = Rz*X3
     X4_flat = P*X4
     graph01 += line(X4_flat.transpose(),color="blue")
     show(graph01,aspect_ratio=1,axes=False)

     
     
135. 
     

     Question

     prevnext

     One of the simplest 3D \(\to\) 2D projections (to draw by hand) is an oblique projection. Each type of oblique projection can be parameterized with an angle (\(\theta\)) and a ratio (\(r<1\)). Below is an oblique-projection matrix.

     \[P = \begin{bmatrix}1 & r\cos(\theta_P) & 0 \\ 0 & r\sin(\theta_P) & 1 \end{bmatrix}\]

     We can exemplify an oblique projection with \(r=0.5\) and \(\theta_P=\frac{pi}{6}\) on a unit cube. The cube has lengths 1 unit long.

     (SageMath)

     X1 = matrix([[0,0,0,0,0,1,1,0,1,1,0,1,1,0,1,1],[0,0,1,1,0,0,1,1,1,1,1,1,0,0,0,0],[0,1,1,0,0,0,0,0,0,1,1,1,1,1,1,0]])
     r = 0.5
     tp = pi/6
     P = matrix([[1,r*cos(tp),0],[0,r*sin(tp),1]])
     flat_X1 = P*X1
     line(flat_X1.transpose(),aspect_ratio=1)

     
     

     In 3D, there are relatively simple rotation matrices for spinning around the \(x\), \(y\), or \(z\) axis.

     \[R_x = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos(\theta_x) & -\sin(\theta_x) \\ 0 &\sin(\theta_x) & \cos(\theta_x) \end{bmatrix}\]

     \[R_y = \begin{bmatrix}\cos(\theta_y) & 0 & \sin(\theta_y) \\ 0 & 1 & 0 \\ -\sin(\theta_y) & 0 & \cos(\theta_y) \end{bmatrix}\]

     \[R_z = \begin{bmatrix}\cos(\theta_z) & -\sin(\theta_z) & 0 \\ \sin(\theta_z) & \cos(\theta_z) & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     Information specific to this question

     Fill in the missing code to generate various rotations of a square pyramid.

     • Red: X1 is the original
     • Orange: X2 is after a rotation of 3.77 radians around the x-axis
     • Green: X3 has another rotation of 3.72 radians around the y-axis
     • Blue: X4 has another rotation of 4.11 radians around the z-axis

     In order to flatten the 3D points onto a 2D graph, each pyramid will undergo an oblique projection with \(r=0.5\) and \(\theta_P=0.71\).

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = ...
     X1_flat = ...
     graph01 += line(X1_flat.transpose(),color="red")
     Rx = ...
     X2 = ...
     X2_flat = ...
     graph01 += line(X2_flat.transpose(),color="orange")
     Ry = ...
     X3 = ...
     X3_flat = ...
     graph01 += ...
     Rz = ...
     X4 = ...
     X4_flat = ...
     graph01 += ...
     show(graph01,aspect_ratio=1,axes=False)

     Which graph matches?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is B.

     SageMath

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = matrix([[1,0.5*cos(0.71),0],[0,0.5*sin(0.71),1]])
     X1_flat = P*X1
     graph01 += line(X1_flat.transpose(),color="red")
     tx = 3.77
     Rx = matrix([[1,0,0],[0,cos(tx),-sin(tx)],[0,sin(tx),cos(tx)]])
     X2 = Rx*X1
     X2_flat = P*X2
     graph01 += line(X2_flat.transpose(),color="orange")
     ty = 3.72
     Ry = matrix([[cos(ty),0,sin(ty)],[0,1,0],[-sin(ty),0,cos(ty)]])
     X3 = Ry*X2
     X3_flat = P*X3
     graph01 += line(X3_flat.transpose(),color="green")
     Rz = matrix([[cos(4.11),-sin(4.11),0],[sin(4.11),cos(4.11),0],[0,0,1]])
     X4 = Rz*X3
     X4_flat = P*X4
     graph01 += line(X4_flat.transpose(),color="blue")
     show(graph01,aspect_ratio=1,axes=False)

     
     
136. 
     

     Question

     prevnext

     One of the simplest 3D \(\to\) 2D projections (to draw by hand) is an oblique projection. Each type of oblique projection can be parameterized with an angle (\(\theta\)) and a ratio (\(r<1\)). Below is an oblique-projection matrix.

     \[P = \begin{bmatrix}1 & r\cos(\theta_P) & 0 \\ 0 & r\sin(\theta_P) & 1 \end{bmatrix}\]

     We can exemplify an oblique projection with \(r=0.5\) and \(\theta_P=\frac{pi}{6}\) on a unit cube. The cube has lengths 1 unit long.

     (SageMath)

     X1 = matrix([[0,0,0,0,0,1,1,0,1,1,0,1,1,0,1,1],[0,0,1,1,0,0,1,1,1,1,1,1,0,0,0,0],[0,1,1,0,0,0,0,0,0,1,1,1,1,1,1,0]])
     r = 0.5
     tp = pi/6
     P = matrix([[1,r*cos(tp),0],[0,r*sin(tp),1]])
     flat_X1 = P*X1
     line(flat_X1.transpose(),aspect_ratio=1)

     
     

     In 3D, there are relatively simple rotation matrices for spinning around the \(x\), \(y\), or \(z\) axis.

     \[R_x = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos(\theta_x) & -\sin(\theta_x) \\ 0 &\sin(\theta_x) & \cos(\theta_x) \end{bmatrix}\]

     \[R_y = \begin{bmatrix}\cos(\theta_y) & 0 & \sin(\theta_y) \\ 0 & 1 & 0 \\ -\sin(\theta_y) & 0 & \cos(\theta_y) \end{bmatrix}\]

     \[R_z = \begin{bmatrix}\cos(\theta_z) & -\sin(\theta_z) & 0 \\ \sin(\theta_z) & \cos(\theta_z) & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     Information specific to this question

     Fill in the missing code to generate various rotations of a square pyramid.

     • Red: X1 is the original
     • Orange: X2 is after a rotation of 5.59 radians around the x-axis
     • Green: X3 has another rotation of 1.26 radians around the y-axis
     • Blue: X4 has another rotation of 1.1 radians around the z-axis

     In order to flatten the 3D points onto a 2D graph, each pyramid will undergo an oblique projection with \(r=0.5\) and \(\theta_P=0.67\).

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = ...
     X1_flat = ...
     graph01 += line(X1_flat.transpose(),color="red")
     Rx = ...
     X2 = ...
     X2_flat = ...
     graph01 += line(X2_flat.transpose(),color="orange")
     Ry = ...
     X3 = ...
     X3_flat = ...
     graph01 += ...
     Rz = ...
     X4 = ...
     X4_flat = ...
     graph01 += ...
     show(graph01,aspect_ratio=1,axes=False)

     Which graph matches?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is A.

     SageMath

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = matrix([[1,0.5*cos(0.67),0],[0,0.5*sin(0.67),1]])
     X1_flat = P*X1
     graph01 += line(X1_flat.transpose(),color="red")
     tx = 5.59
     Rx = matrix([[1,0,0],[0,cos(tx),-sin(tx)],[0,sin(tx),cos(tx)]])
     X2 = Rx*X1
     X2_flat = P*X2
     graph01 += line(X2_flat.transpose(),color="orange")
     ty = 1.26
     Ry = matrix([[cos(ty),0,sin(ty)],[0,1,0],[-sin(ty),0,cos(ty)]])
     X3 = Ry*X2
     X3_flat = P*X3
     graph01 += line(X3_flat.transpose(),color="green")
     Rz = matrix([[cos(1.1),-sin(1.1),0],[sin(1.1),cos(1.1),0],[0,0,1]])
     X4 = Rz*X3
     X4_flat = P*X4
     graph01 += line(X4_flat.transpose(),color="blue")
     show(graph01,aspect_ratio=1,axes=False)

     
     
137. 
     

     Question

     prevnext

     One of the simplest 3D \(\to\) 2D projections (to draw by hand) is an oblique projection. Each type of oblique projection can be parameterized with an angle (\(\theta\)) and a ratio (\(r<1\)). Below is an oblique-projection matrix.

     \[P = \begin{bmatrix}1 & r\cos(\theta_P) & 0 \\ 0 & r\sin(\theta_P) & 1 \end{bmatrix}\]

     We can exemplify an oblique projection with \(r=0.5\) and \(\theta_P=\frac{pi}{6}\) on a unit cube. The cube has lengths 1 unit long.

     (SageMath)

     X1 = matrix([[0,0,0,0,0,1,1,0,1,1,0,1,1,0,1,1],[0,0,1,1,0,0,1,1,1,1,1,1,0,0,0,0],[0,1,1,0,0,0,0,0,0,1,1,1,1,1,1,0]])
     r = 0.5
     tp = pi/6
     P = matrix([[1,r*cos(tp),0],[0,r*sin(tp),1]])
     flat_X1 = P*X1
     line(flat_X1.transpose(),aspect_ratio=1)

     
     

     In 3D, there are relatively simple rotation matrices for spinning around the \(x\), \(y\), or \(z\) axis.

     \[R_x = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos(\theta_x) & -\sin(\theta_x) \\ 0 &\sin(\theta_x) & \cos(\theta_x) \end{bmatrix}\]

     \[R_y = \begin{bmatrix}\cos(\theta_y) & 0 & \sin(\theta_y) \\ 0 & 1 & 0 \\ -\sin(\theta_y) & 0 & \cos(\theta_y) \end{bmatrix}\]

     \[R_z = \begin{bmatrix}\cos(\theta_z) & -\sin(\theta_z) & 0 \\ \sin(\theta_z) & \cos(\theta_z) & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     Information specific to this question

     Fill in the missing code to generate various rotations of a square pyramid.

     • Red: X1 is the original
     • Orange: X2 is after a rotation of 0.67 radians around the x-axis
     • Green: X3 has another rotation of 2.31 radians around the y-axis
     • Blue: X4 has another rotation of 6.12 radians around the z-axis

     In order to flatten the 3D points onto a 2D graph, each pyramid will undergo an oblique projection with \(r=0.5\) and \(\theta_P=0.71\).

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = ...
     X1_flat = ...
     graph01 += line(X1_flat.transpose(),color="red")
     Rx = ...
     X2 = ...
     X2_flat = ...
     graph01 += line(X2_flat.transpose(),color="orange")
     Ry = ...
     X3 = ...
     X3_flat = ...
     graph01 += ...
     Rz = ...
     X4 = ...
     X4_flat = ...
     graph01 += ...
     show(graph01,aspect_ratio=1,axes=False)

     Which graph matches?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is B.

     SageMath

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = matrix([[1,0.5*cos(0.71),0],[0,0.5*sin(0.71),1]])
     X1_flat = P*X1
     graph01 += line(X1_flat.transpose(),color="red")
     tx = 0.67
     Rx = matrix([[1,0,0],[0,cos(tx),-sin(tx)],[0,sin(tx),cos(tx)]])
     X2 = Rx*X1
     X2_flat = P*X2
     graph01 += line(X2_flat.transpose(),color="orange")
     ty = 2.31
     Ry = matrix([[cos(ty),0,sin(ty)],[0,1,0],[-sin(ty),0,cos(ty)]])
     X3 = Ry*X2
     X3_flat = P*X3
     graph01 += line(X3_flat.transpose(),color="green")
     Rz = matrix([[cos(6.12),-sin(6.12),0],[sin(6.12),cos(6.12),0],[0,0,1]])
     X4 = Rz*X3
     X4_flat = P*X4
     graph01 += line(X4_flat.transpose(),color="blue")
     show(graph01,aspect_ratio=1,axes=False)

     
     
138. 
     

     Question

     prevnext

     One of the simplest 3D \(\to\) 2D projections (to draw by hand) is an oblique projection. Each type of oblique projection can be parameterized with an angle (\(\theta\)) and a ratio (\(r<1\)). Below is an oblique-projection matrix.

     \[P = \begin{bmatrix}1 & r\cos(\theta_P) & 0 \\ 0 & r\sin(\theta_P) & 1 \end{bmatrix}\]

     We can exemplify an oblique projection with \(r=0.5\) and \(\theta_P=\frac{pi}{6}\) on a unit cube. The cube has lengths 1 unit long.

     (SageMath)

     X1 = matrix([[0,0,0,0,0,1,1,0,1,1,0,1,1,0,1,1],[0,0,1,1,0,0,1,1,1,1,1,1,0,0,0,0],[0,1,1,0,0,0,0,0,0,1,1,1,1,1,1,0]])
     r = 0.5
     tp = pi/6
     P = matrix([[1,r*cos(tp),0],[0,r*sin(tp),1]])
     flat_X1 = P*X1
     line(flat_X1.transpose(),aspect_ratio=1)

     
     

     In 3D, there are relatively simple rotation matrices for spinning around the \(x\), \(y\), or \(z\) axis.

     \[R_x = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos(\theta_x) & -\sin(\theta_x) \\ 0 &\sin(\theta_x) & \cos(\theta_x) \end{bmatrix}\]

     \[R_y = \begin{bmatrix}\cos(\theta_y) & 0 & \sin(\theta_y) \\ 0 & 1 & 0 \\ -\sin(\theta_y) & 0 & \cos(\theta_y) \end{bmatrix}\]

     \[R_z = \begin{bmatrix}\cos(\theta_z) & -\sin(\theta_z) & 0 \\ \sin(\theta_z) & \cos(\theta_z) & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     Information specific to this question

     Fill in the missing code to generate various rotations of a square pyramid.

     • Red: X1 is the original
     • Orange: X2 is after a rotation of 6.15 radians around the x-axis
     • Green: X3 has another rotation of 0.71 radians around the y-axis
     • Blue: X4 has another rotation of 5.55 radians around the z-axis

     In order to flatten the 3D points onto a 2D graph, each pyramid will undergo an oblique projection with \(r=0.5\) and \(\theta_P=0.25\).

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = ...
     X1_flat = ...
     graph01 += line(X1_flat.transpose(),color="red")
     Rx = ...
     X2 = ...
     X2_flat = ...
     graph01 += line(X2_flat.transpose(),color="orange")
     Ry = ...
     X3 = ...
     X3_flat = ...
     graph01 += ...
     Rz = ...
     X4 = ...
     X4_flat = ...
     graph01 += ...
     show(graph01,aspect_ratio=1,axes=False)

     Which graph matches?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is B.

     SageMath

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = matrix([[1,0.5*cos(0.25),0],[0,0.5*sin(0.25),1]])
     X1_flat = P*X1
     graph01 += line(X1_flat.transpose(),color="red")
     tx = 6.15
     Rx = matrix([[1,0,0],[0,cos(tx),-sin(tx)],[0,sin(tx),cos(tx)]])
     X2 = Rx*X1
     X2_flat = P*X2
     graph01 += line(X2_flat.transpose(),color="orange")
     ty = 0.71
     Ry = matrix([[cos(ty),0,sin(ty)],[0,1,0],[-sin(ty),0,cos(ty)]])
     X3 = Ry*X2
     X3_flat = P*X3
     graph01 += line(X3_flat.transpose(),color="green")
     Rz = matrix([[cos(5.55),-sin(5.55),0],[sin(5.55),cos(5.55),0],[0,0,1]])
     X4 = Rz*X3
     X4_flat = P*X4
     graph01 += line(X4_flat.transpose(),color="blue")
     show(graph01,aspect_ratio=1,axes=False)

     
     
139. 
     

     Question

     prevnext

     One of the simplest 3D \(\to\) 2D projections (to draw by hand) is an oblique projection. Each type of oblique projection can be parameterized with an angle (\(\theta\)) and a ratio (\(r<1\)). Below is an oblique-projection matrix.

     \[P = \begin{bmatrix}1 & r\cos(\theta_P) & 0 \\ 0 & r\sin(\theta_P) & 1 \end{bmatrix}\]

     We can exemplify an oblique projection with \(r=0.5\) and \(\theta_P=\frac{pi}{6}\) on a unit cube. The cube has lengths 1 unit long.

     (SageMath)

     X1 = matrix([[0,0,0,0,0,1,1,0,1,1,0,1,1,0,1,1],[0,0,1,1,0,0,1,1,1,1,1,1,0,0,0,0],[0,1,1,0,0,0,0,0,0,1,1,1,1,1,1,0]])
     r = 0.5
     tp = pi/6
     P = matrix([[1,r*cos(tp),0],[0,r*sin(tp),1]])
     flat_X1 = P*X1
     line(flat_X1.transpose(),aspect_ratio=1)

     
     

     In 3D, there are relatively simple rotation matrices for spinning around the \(x\), \(y\), or \(z\) axis.

     \[R_x = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos(\theta_x) & -\sin(\theta_x) \\ 0 &\sin(\theta_x) & \cos(\theta_x) \end{bmatrix}\]

     \[R_y = \begin{bmatrix}\cos(\theta_y) & 0 & \sin(\theta_y) \\ 0 & 1 & 0 \\ -\sin(\theta_y) & 0 & \cos(\theta_y) \end{bmatrix}\]

     \[R_z = \begin{bmatrix}\cos(\theta_z) & -\sin(\theta_z) & 0 \\ \sin(\theta_z) & \cos(\theta_z) & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     Information specific to this question

     Fill in the missing code to generate various rotations of a square pyramid.

     • Red: X1 is the original
     • Orange: X2 is after a rotation of 4.19 radians around the x-axis
     • Green: X3 has another rotation of 1.91 radians around the y-axis
     • Blue: X4 has another rotation of 2.43 radians around the z-axis

     In order to flatten the 3D points onto a 2D graph, each pyramid will undergo an oblique projection with \(r=0.5\) and \(\theta_P=0.68\).

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = ...
     X1_flat = ...
     graph01 += line(X1_flat.transpose(),color="red")
     Rx = ...
     X2 = ...
     X2_flat = ...
     graph01 += line(X2_flat.transpose(),color="orange")
     Ry = ...
     X3 = ...
     X3_flat = ...
     graph01 += ...
     Rz = ...
     X4 = ...
     X4_flat = ...
     graph01 += ...
     show(graph01,aspect_ratio=1,axes=False)

     Which graph matches?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is A.

     SageMath

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = matrix([[1,0.5*cos(0.68),0],[0,0.5*sin(0.68),1]])
     X1_flat = P*X1
     graph01 += line(X1_flat.transpose(),color="red")
     tx = 4.19
     Rx = matrix([[1,0,0],[0,cos(tx),-sin(tx)],[0,sin(tx),cos(tx)]])
     X2 = Rx*X1
     X2_flat = P*X2
     graph01 += line(X2_flat.transpose(),color="orange")
     ty = 1.91
     Ry = matrix([[cos(ty),0,sin(ty)],[0,1,0],[-sin(ty),0,cos(ty)]])
     X3 = Ry*X2
     X3_flat = P*X3
     graph01 += line(X3_flat.transpose(),color="green")
     Rz = matrix([[cos(2.43),-sin(2.43),0],[sin(2.43),cos(2.43),0],[0,0,1]])
     X4 = Rz*X3
     X4_flat = P*X4
     graph01 += line(X4_flat.transpose(),color="blue")
     show(graph01,aspect_ratio=1,axes=False)

     
     
140. 
     

     Question

     prevnext

     One of the simplest 3D \(\to\) 2D projections (to draw by hand) is an oblique projection. Each type of oblique projection can be parameterized with an angle (\(\theta\)) and a ratio (\(r<1\)). Below is an oblique-projection matrix.

     \[P = \begin{bmatrix}1 & r\cos(\theta_P) & 0 \\ 0 & r\sin(\theta_P) & 1 \end{bmatrix}\]

     We can exemplify an oblique projection with \(r=0.5\) and \(\theta_P=\frac{pi}{6}\) on a unit cube. The cube has lengths 1 unit long.

     (SageMath)

     X1 = matrix([[0,0,0,0,0,1,1,0,1,1,0,1,1,0,1,1],[0,0,1,1,0,0,1,1,1,1,1,1,0,0,0,0],[0,1,1,0,0,0,0,0,0,1,1,1,1,1,1,0]])
     r = 0.5
     tp = pi/6
     P = matrix([[1,r*cos(tp),0],[0,r*sin(tp),1]])
     flat_X1 = P*X1
     line(flat_X1.transpose(),aspect_ratio=1)

     
     

     In 3D, there are relatively simple rotation matrices for spinning around the \(x\), \(y\), or \(z\) axis.

     \[R_x = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos(\theta_x) & -\sin(\theta_x) \\ 0 &\sin(\theta_x) & \cos(\theta_x) \end{bmatrix}\]

     \[R_y = \begin{bmatrix}\cos(\theta_y) & 0 & \sin(\theta_y) \\ 0 & 1 & 0 \\ -\sin(\theta_y) & 0 & \cos(\theta_y) \end{bmatrix}\]

     \[R_z = \begin{bmatrix}\cos(\theta_z) & -\sin(\theta_z) & 0 \\ \sin(\theta_z) & \cos(\theta_z) & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     Information specific to this question

     Fill in the missing code to generate various rotations of a square pyramid.

     • Red: X1 is the original
     • Orange: X2 is after a rotation of 5.65 radians around the x-axis
     • Green: X3 has another rotation of 3.85 radians around the y-axis
     • Blue: X4 has another rotation of 5.12 radians around the z-axis

     In order to flatten the 3D points onto a 2D graph, each pyramid will undergo an oblique projection with \(r=0.5\) and \(\theta_P=0.65\).

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = ...
     X1_flat = ...
     graph01 += line(X1_flat.transpose(),color="red")
     Rx = ...
     X2 = ...
     X2_flat = ...
     graph01 += line(X2_flat.transpose(),color="orange")
     Ry = ...
     X3 = ...
     X3_flat = ...
     graph01 += ...
     Rz = ...
     X4 = ...
     X4_flat = ...
     graph01 += ...
     show(graph01,aspect_ratio=1,axes=False)

     Which graph matches?

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is B.

     SageMath

     graph01 = Graphics()
     X1 = matrix([[1,-1,-1,1,1,0,1,-1,0,-1],[-1,-1,1,1,-1,0,1,1,0,-1],[-1,-1,-1,-1,-1,1,-1,-1,1,-1]])
     P = matrix([[1,0.5*cos(0.65),0],[0,0.5*sin(0.65),1]])
     X1_flat = P*X1
     graph01 += line(X1_flat.transpose(),color="red")
     tx = 5.65
     Rx = matrix([[1,0,0],[0,cos(tx),-sin(tx)],[0,sin(tx),cos(tx)]])
     X2 = Rx*X1
     X2_flat = P*X2
     graph01 += line(X2_flat.transpose(),color="orange")
     ty = 3.85
     Ry = matrix([[cos(ty),0,sin(ty)],[0,1,0],[-sin(ty),0,cos(ty)]])
     X3 = Ry*X2
     X3_flat = P*X3
     graph01 += line(X3_flat.transpose(),color="green")
     Rz = matrix([[cos(5.12),-sin(5.12),0],[sin(5.12),cos(5.12),0],[0,0,1]])
     X4 = Rz*X3
     X4_flat = P*X4
     graph01 += line(X4_flat.transpose(),color="blue")
     show(graph01,aspect_ratio=1,axes=False)

     
     
141. 
     

     Question

     prevnext

     Your goal is to make an animation (with 20 frames) of square pyramid spinning around a given axis. In 3D space, direction can be indicated with a unit vector \(\vec{u}\):

     \[\vec{u} = \begin{bmatrix}u_x \\ u_y \\ u_z \end{bmatrix}\] A unit vector has a length of 1, so \(u_x^2+u_y^2+u_z^2=1\).

     In order to rotate points around the axis \(\vec{u}\) by an angle \(\theta\), we can use the general 3D rotation matrix:

     \[R = \begin{bmatrix} \cos(\theta)+u_x^2(1-\cos(\theta)) & u_xu_y(1-\cos(\theta))-u_z\sin(\theta) & u_xu_z(1-\cos(\theta))+u_y\sin(\theta) \\ u_yu_x(1-\cos(\theta))+u_z\sin(\theta) & \cos(\theta)+u_y^2(1-\cos(\theta)) & u_yu_z(1-\cos(\theta))-u_x\sin(\theta) \\ u_zu_x(1-\cos(\theta))-u_y\sin(\theta) & u_zu_y(1-\cos(\theta))+u_x\sin(\theta) & \cos(\theta)+u_z^2(1-\cos(\theta)) \end{bmatrix}\] Because our animation has 20 frames, the angle should be \(\theta=\frac{2\pi}{20}\).

     We will use the basic orthographic projection matrix below to flatten 3D points into a 2D plane:

     \[P = \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     The axis of rotation should be:

     \[u_x=0.3259576\] \[u_y=0.3402856\] \[u_z=-0.8820189\]

     Finish the code below to graph the animation.

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = ######### ENTER YOUR CODE HERE ##############
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     A

     
     

     B

     
     

     C

     
     

     D

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is D.

     SageMath

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = matrix([[0.956256681196475,0.27798757925661,0.0910827395621813],[-0.267130086564853,0.956723891743267,-0.115416254572186],[-0.119225318279955,0.0860366244495171,0.989132459650565]])
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     Instead of calculating each element of the rotation matrix (R), I would recommend putting in the formulas in terms of \(u_x\), \(u_y\), \(u_z\), and \(\theta\). You would need to define those variables before they were referenced:

     ux = 0.3259576
     uy = 0.3402856
     uz = -0.8820189
     th = 2*pi/20

     
     
142. 
     

     Question

     prevnext

     Your goal is to make an animation (with 20 frames) of square pyramid spinning around a given axis. In 3D space, direction can be indicated with a unit vector \(\vec{u}\):

     \[\vec{u} = \begin{bmatrix}u_x \\ u_y \\ u_z \end{bmatrix}\] A unit vector has a length of 1, so \(u_x^2+u_y^2+u_z^2=1\).

     In order to rotate points around the axis \(\vec{u}\) by an angle \(\theta\), we can use the general 3D rotation matrix:

     \[R = \begin{bmatrix} \cos(\theta)+u_x^2(1-\cos(\theta)) & u_xu_y(1-\cos(\theta))-u_z\sin(\theta) & u_xu_z(1-\cos(\theta))+u_y\sin(\theta) \\ u_yu_x(1-\cos(\theta))+u_z\sin(\theta) & \cos(\theta)+u_y^2(1-\cos(\theta)) & u_yu_z(1-\cos(\theta))-u_x\sin(\theta) \\ u_zu_x(1-\cos(\theta))-u_y\sin(\theta) & u_zu_y(1-\cos(\theta))+u_x\sin(\theta) & \cos(\theta)+u_z^2(1-\cos(\theta)) \end{bmatrix}\] Because our animation has 20 frames, the angle should be \(\theta=\frac{2\pi}{20}\).

     We will use the basic orthographic projection matrix below to flatten 3D points into a 2D plane:

     \[P = \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     The axis of rotation should be:

     \[u_x=0.8137197\] \[u_y=0.2912837\] \[u_z=0.5030051\]

     Finish the code below to graph the animation.

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = ######### ENTER YOUR CODE HERE ##############
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     A

     
     

     B

     
     

     C

     
     

     D

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is A.

     SageMath

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = matrix([[0.983463939757809,-0.143836376557919,0.110044427277067],[0.167037864820034,0.95520918411058,-0.244282144879981],[-0.0699787890179099,0.258624266783859,0.963439908721918]])
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     Instead of calculating each element of the rotation matrix (R), I would recommend putting in the formulas in terms of \(u_x\), \(u_y\), \(u_z\), and \(\theta\). You would need to define those variables before they were referenced:

     ux = 0.8137197
     uy = 0.2912837
     uz = 0.5030051
     th = 2*pi/20

     
     
143. 
     

     Question

     prevnext

     Your goal is to make an animation (with 20 frames) of square pyramid spinning around a given axis. In 3D space, direction can be indicated with a unit vector \(\vec{u}\):

     \[\vec{u} = \begin{bmatrix}u_x \\ u_y \\ u_z \end{bmatrix}\] A unit vector has a length of 1, so \(u_x^2+u_y^2+u_z^2=1\).

     In order to rotate points around the axis \(\vec{u}\) by an angle \(\theta\), we can use the general 3D rotation matrix:

     \[R = \begin{bmatrix} \cos(\theta)+u_x^2(1-\cos(\theta)) & u_xu_y(1-\cos(\theta))-u_z\sin(\theta) & u_xu_z(1-\cos(\theta))+u_y\sin(\theta) \\ u_yu_x(1-\cos(\theta))+u_z\sin(\theta) & \cos(\theta)+u_y^2(1-\cos(\theta)) & u_yu_z(1-\cos(\theta))-u_x\sin(\theta) \\ u_zu_x(1-\cos(\theta))-u_y\sin(\theta) & u_zu_y(1-\cos(\theta))+u_x\sin(\theta) & \cos(\theta)+u_z^2(1-\cos(\theta)) \end{bmatrix}\] Because our animation has 20 frames, the angle should be \(\theta=\frac{2\pi}{20}\).

     We will use the basic orthographic projection matrix below to flatten 3D points into a 2D plane:

     \[P = \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     The axis of rotation should be:

     \[u_x=-0.6178855\] \[u_y=0.0214379\] \[u_z=-0.7859758\]

     Finish the code below to graph the animation.

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = ######### ENTER YOUR CODE HERE ##############
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     A

     
     

     B

     
     

     C

     
     

     D

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is C.

     SageMath

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = matrix([[0.969742279866997,0.242231566713181,0.0303937283997408],[-0.243528192118999,0.951079009856047,0.1901124316148],[0.0171443950494465,-0.19176179259414,0.981291742867263]])
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     Instead of calculating each element of the rotation matrix (R), I would recommend putting in the formulas in terms of \(u_x\), \(u_y\), \(u_z\), and \(\theta\). You would need to define those variables before they were referenced:

     ux = -0.6178855
     uy = 0.0214379
     uz = -0.7859758
     th = 2*pi/20

     
     
144. 
     

     Question

     prevnext

     Your goal is to make an animation (with 20 frames) of square pyramid spinning around a given axis. In 3D space, direction can be indicated with a unit vector \(\vec{u}\):

     \[\vec{u} = \begin{bmatrix}u_x \\ u_y \\ u_z \end{bmatrix}\] A unit vector has a length of 1, so \(u_x^2+u_y^2+u_z^2=1\).

     In order to rotate points around the axis \(\vec{u}\) by an angle \(\theta\), we can use the general 3D rotation matrix:

     \[R = \begin{bmatrix} \cos(\theta)+u_x^2(1-\cos(\theta)) & u_xu_y(1-\cos(\theta))-u_z\sin(\theta) & u_xu_z(1-\cos(\theta))+u_y\sin(\theta) \\ u_yu_x(1-\cos(\theta))+u_z\sin(\theta) & \cos(\theta)+u_y^2(1-\cos(\theta)) & u_yu_z(1-\cos(\theta))-u_x\sin(\theta) \\ u_zu_x(1-\cos(\theta))-u_y\sin(\theta) & u_zu_y(1-\cos(\theta))+u_x\sin(\theta) & \cos(\theta)+u_z^2(1-\cos(\theta)) \end{bmatrix}\] Because our animation has 20 frames, the angle should be \(\theta=\frac{2\pi}{20}\).

     We will use the basic orthographic projection matrix below to flatten 3D points into a 2D plane:

     \[P = \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     The axis of rotation should be:

     \[u_x=0.3080886\] \[u_y=0.9339919\] \[u_z=-0.1809436\]

     Finish the code below to graph the animation.

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = ######### ENTER YOUR CODE HERE ##############
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     A

     
     

     B

     
     

     C

     
     

     D

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is A.

     SageMath

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = matrix([[0.955702162146452,0.0699982477474729,0.285890927769455],[-0.0418310537330161,0.993751915602564,-0.103476051238051],[-0.291347799394781,0.0869331671372767,0.952658954841291]])
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     Instead of calculating each element of the rotation matrix (R), I would recommend putting in the formulas in terms of \(u_x\), \(u_y\), \(u_z\), and \(\theta\). You would need to define those variables before they were referenced:

     ux = 0.3080886
     uy = 0.9339919
     uz = -0.1809436
     th = 2*pi/20

     
     
145. 
     

     Question

     prevnext

     Your goal is to make an animation (with 20 frames) of square pyramid spinning around a given axis. In 3D space, direction can be indicated with a unit vector \(\vec{u}\):

     \[\vec{u} = \begin{bmatrix}u_x \\ u_y \\ u_z \end{bmatrix}\] A unit vector has a length of 1, so \(u_x^2+u_y^2+u_z^2=1\).

     In order to rotate points around the axis \(\vec{u}\) by an angle \(\theta\), we can use the general 3D rotation matrix:

     \[R = \begin{bmatrix} \cos(\theta)+u_x^2(1-\cos(\theta)) & u_xu_y(1-\cos(\theta))-u_z\sin(\theta) & u_xu_z(1-\cos(\theta))+u_y\sin(\theta) \\ u_yu_x(1-\cos(\theta))+u_z\sin(\theta) & \cos(\theta)+u_y^2(1-\cos(\theta)) & u_yu_z(1-\cos(\theta))-u_x\sin(\theta) \\ u_zu_x(1-\cos(\theta))-u_y\sin(\theta) & u_zu_y(1-\cos(\theta))+u_x\sin(\theta) & \cos(\theta)+u_z^2(1-\cos(\theta)) \end{bmatrix}\] Because our animation has 20 frames, the angle should be \(\theta=\frac{2\pi}{20}\).

     We will use the basic orthographic projection matrix below to flatten 3D points into a 2D plane:

     \[P = \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     The axis of rotation should be:

     \[u_x=-0.7237868\] \[u_y=0.6642399\] \[u_z=-0.1868634\]

     Finish the code below to graph the animation.

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = ######### ENTER YOUR CODE HERE ##############
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     A

     
     

     B

     
     

     C

     
     

     D

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is D.

     SageMath

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = matrix([[0.976696411476126,0.034213492779094,0.211880996603075],[-0.0812744262320707,0.9726511000662,0.217587465587194],[-0.198641857246414,-0.229737403249591,0.95276552104798]])
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     Instead of calculating each element of the rotation matrix (R), I would recommend putting in the formulas in terms of \(u_x\), \(u_y\), \(u_z\), and \(\theta\). You would need to define those variables before they were referenced:

     ux = -0.7237868
     uy = 0.6642399
     uz = -0.1868634
     th = 2*pi/20

     
     
146. 
     

     Question

     prevnext

     Your goal is to make an animation (with 20 frames) of square pyramid spinning around a given axis. In 3D space, direction can be indicated with a unit vector \(\vec{u}\):

     \[\vec{u} = \begin{bmatrix}u_x \\ u_y \\ u_z \end{bmatrix}\] A unit vector has a length of 1, so \(u_x^2+u_y^2+u_z^2=1\).

     In order to rotate points around the axis \(\vec{u}\) by an angle \(\theta\), we can use the general 3D rotation matrix:

     \[R = \begin{bmatrix} \cos(\theta)+u_x^2(1-\cos(\theta)) & u_xu_y(1-\cos(\theta))-u_z\sin(\theta) & u_xu_z(1-\cos(\theta))+u_y\sin(\theta) \\ u_yu_x(1-\cos(\theta))+u_z\sin(\theta) & \cos(\theta)+u_y^2(1-\cos(\theta)) & u_yu_z(1-\cos(\theta))-u_x\sin(\theta) \\ u_zu_x(1-\cos(\theta))-u_y\sin(\theta) & u_zu_y(1-\cos(\theta))+u_x\sin(\theta) & \cos(\theta)+u_z^2(1-\cos(\theta)) \end{bmatrix}\] Because our animation has 20 frames, the angle should be \(\theta=\frac{2\pi}{20}\).

     We will use the basic orthographic projection matrix below to flatten 3D points into a 2D plane:

     \[P = \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     The axis of rotation should be:

     \[u_x=0.0684043\] \[u_y=0.3809468\] \[u_z=-0.9220631\]

     Finish the code below to graph the animation.

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = ######### ENTER YOUR CODE HERE ##############
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     A

     
     

     B

     
     

     C

     
     

     D

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is C.

     SageMath

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = matrix([[0.951285530251319,0.286208561141078,0.114632017621694],[-0.283657782696272,0.958159217352581,-0.0383298384828918],[-0.120806052208908,0.00394635678107557,0.992668284986407]])
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     Instead of calculating each element of the rotation matrix (R), I would recommend putting in the formulas in terms of \(u_x\), \(u_y\), \(u_z\), and \(\theta\). You would need to define those variables before they were referenced:

     ux = 0.0684043
     uy = 0.3809468
     uz = -0.9220631
     th = 2*pi/20

     
     
147. 
     

     Question

     prevnext

     Your goal is to make an animation (with 20 frames) of square pyramid spinning around a given axis. In 3D space, direction can be indicated with a unit vector \(\vec{u}\):

     \[\vec{u} = \begin{bmatrix}u_x \\ u_y \\ u_z \end{bmatrix}\] A unit vector has a length of 1, so \(u_x^2+u_y^2+u_z^2=1\).

     In order to rotate points around the axis \(\vec{u}\) by an angle \(\theta\), we can use the general 3D rotation matrix:

     \[R = \begin{bmatrix} \cos(\theta)+u_x^2(1-\cos(\theta)) & u_xu_y(1-\cos(\theta))-u_z\sin(\theta) & u_xu_z(1-\cos(\theta))+u_y\sin(\theta) \\ u_yu_x(1-\cos(\theta))+u_z\sin(\theta) & \cos(\theta)+u_y^2(1-\cos(\theta)) & u_yu_z(1-\cos(\theta))-u_x\sin(\theta) \\ u_zu_x(1-\cos(\theta))-u_y\sin(\theta) & u_zu_y(1-\cos(\theta))+u_x\sin(\theta) & \cos(\theta)+u_z^2(1-\cos(\theta)) \end{bmatrix}\] Because our animation has 20 frames, the angle should be \(\theta=\frac{2\pi}{20}\).

     We will use the basic orthographic projection matrix below to flatten 3D points into a 2D plane:

     \[P = \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     The axis of rotation should be:

     \[u_x=-0.3834826\] \[u_y=0.2169223\] \[u_z=0.8977114\]

     Finish the code below to graph the animation.

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = ######### ENTER YOUR CODE HERE ##############
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     A

     
     

     B

     
     

     C

     
     

     D

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is D.

     SageMath

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = matrix([[0.958254090771837,-0.281479497040066,0.0501835656875039],[0.27333667791739,0.953359566561581,0.12803357899629],[-0.0838818098524688,-0.108971691698283,0.99049937525689]])
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     Instead of calculating each element of the rotation matrix (R), I would recommend putting in the formulas in terms of \(u_x\), \(u_y\), \(u_z\), and \(\theta\). You would need to define those variables before they were referenced:

     ux = -0.3834826
     uy = 0.2169223
     uz = 0.8977114
     th = 2*pi/20

     
     
148. 
     

     Question

     prevnext

     Your goal is to make an animation (with 20 frames) of square pyramid spinning around a given axis. In 3D space, direction can be indicated with a unit vector \(\vec{u}\):

     \[\vec{u} = \begin{bmatrix}u_x \\ u_y \\ u_z \end{bmatrix}\] A unit vector has a length of 1, so \(u_x^2+u_y^2+u_z^2=1\).

     In order to rotate points around the axis \(\vec{u}\) by an angle \(\theta\), we can use the general 3D rotation matrix:

     \[R = \begin{bmatrix} \cos(\theta)+u_x^2(1-\cos(\theta)) & u_xu_y(1-\cos(\theta))-u_z\sin(\theta) & u_xu_z(1-\cos(\theta))+u_y\sin(\theta) \\ u_yu_x(1-\cos(\theta))+u_z\sin(\theta) & \cos(\theta)+u_y^2(1-\cos(\theta)) & u_yu_z(1-\cos(\theta))-u_x\sin(\theta) \\ u_zu_x(1-\cos(\theta))-u_y\sin(\theta) & u_zu_y(1-\cos(\theta))+u_x\sin(\theta) & \cos(\theta)+u_z^2(1-\cos(\theta)) \end{bmatrix}\] Because our animation has 20 frames, the angle should be \(\theta=\frac{2\pi}{20}\).

     We will use the basic orthographic projection matrix below to flatten 3D points into a 2D plane:

     \[P = \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     The axis of rotation should be:

     \[u_x=0.3024846\] \[u_y=0.9351292\] \[u_z=0.1844896\]

     Finish the code below to graph the animation.

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = ######### ENTER YOUR CODE HERE ##############
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     A

     
     

     B

     
     

     C

     
     

     D

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is A.

     SageMath

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = matrix([[0.955534694624625,-0.0431661466381874,0.291702127440022],[0.0708546677181469,0.993855961917377,-0.0850290716392749],[-0.286239521085358,0.101916685315489,0.952722376048305]])
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     Instead of calculating each element of the rotation matrix (R), I would recommend putting in the formulas in terms of \(u_x\), \(u_y\), \(u_z\), and \(\theta\). You would need to define those variables before they were referenced:

     ux = 0.3024846
     uy = 0.9351292
     uz = 0.1844896
     th = 2*pi/20

     
     
149. 
     

     Question

     prevnext

     Your goal is to make an animation (with 20 frames) of square pyramid spinning around a given axis. In 3D space, direction can be indicated with a unit vector \(\vec{u}\):

     \[\vec{u} = \begin{bmatrix}u_x \\ u_y \\ u_z \end{bmatrix}\] A unit vector has a length of 1, so \(u_x^2+u_y^2+u_z^2=1\).

     In order to rotate points around the axis \(\vec{u}\) by an angle \(\theta\), we can use the general 3D rotation matrix:

     \[R = \begin{bmatrix} \cos(\theta)+u_x^2(1-\cos(\theta)) & u_xu_y(1-\cos(\theta))-u_z\sin(\theta) & u_xu_z(1-\cos(\theta))+u_y\sin(\theta) \\ u_yu_x(1-\cos(\theta))+u_z\sin(\theta) & \cos(\theta)+u_y^2(1-\cos(\theta)) & u_yu_z(1-\cos(\theta))-u_x\sin(\theta) \\ u_zu_x(1-\cos(\theta))-u_y\sin(\theta) & u_zu_y(1-\cos(\theta))+u_x\sin(\theta) & \cos(\theta)+u_z^2(1-\cos(\theta)) \end{bmatrix}\] Because our animation has 20 frames, the angle should be \(\theta=\frac{2\pi}{20}\).

     We will use the basic orthographic projection matrix below to flatten 3D points into a 2D plane:

     \[P = \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     The axis of rotation should be:

     \[u_x=0.7844959\] \[u_y=0.6080293\] \[u_z=-0.1219286\]

     Finish the code below to graph the animation.

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = ######### ENTER YOUR CODE HERE ##############
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     A

     
     

     B

     
     

     C

     
     

     D

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is A.

     SageMath

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = matrix([[0.981177989652168,0.0610238721456914,0.183209824110161],[-0.0143321327006117,0.96915090452915,-0.246051039872996],[-0.192572953953513,0.238794077143209,0.951784138408989]])
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     Instead of calculating each element of the rotation matrix (R), I would recommend putting in the formulas in terms of \(u_x\), \(u_y\), \(u_z\), and \(\theta\). You would need to define those variables before they were referenced:

     ux = 0.7844959
     uy = 0.6080293
     uz = -0.1219286
     th = 2*pi/20

     
     
150. 
     

     Question

     prevnext

     Your goal is to make an animation (with 20 frames) of square pyramid spinning around a given axis. In 3D space, direction can be indicated with a unit vector \(\vec{u}\):

     \[\vec{u} = \begin{bmatrix}u_x \\ u_y \\ u_z \end{bmatrix}\] A unit vector has a length of 1, so \(u_x^2+u_y^2+u_z^2=1\).

     In order to rotate points around the axis \(\vec{u}\) by an angle \(\theta\), we can use the general 3D rotation matrix:

     \[R = \begin{bmatrix} \cos(\theta)+u_x^2(1-\cos(\theta)) & u_xu_y(1-\cos(\theta))-u_z\sin(\theta) & u_xu_z(1-\cos(\theta))+u_y\sin(\theta) \\ u_yu_x(1-\cos(\theta))+u_z\sin(\theta) & \cos(\theta)+u_y^2(1-\cos(\theta)) & u_yu_z(1-\cos(\theta))-u_x\sin(\theta) \\ u_zu_x(1-\cos(\theta))-u_y\sin(\theta) & u_zu_y(1-\cos(\theta))+u_x\sin(\theta) & \cos(\theta)+u_z^2(1-\cos(\theta)) \end{bmatrix}\] Because our animation has 20 frames, the angle should be \(\theta=\frac{2\pi}{20}\).

     We will use the basic orthographic projection matrix below to flatten 3D points into a 2D plane:

     \[P = \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\]

     The axis of rotation should be:

     \[u_x=-0.8137802\] \[u_y=0.5690534\] \[u_z=0.1180675\]

     Finish the code below to graph the animation.

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = ######### ENTER YOUR CODE HERE ##############
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     A

     
     

     B

     
     

     C

     
     

     D

     
     

     
     
     1. A
     2. B
     3. C
     4. D
     
     

     Solution

     Show solution

     The answer is D.

     SageMath

     n = 20 #The number of frames
     X = matrix([[ 1,-1,-1, 1, 1, 0, 1,-1, 0,-1],
                 [-1,-1, 1, 1,-1, 0, 1, 1, 0,-1],
                 [-1,-1,-1,-1,-1, 1,-1,-1, 1,-1]])
     P = matrix([[1,0,0],
                 [0,0,1]])
     R = matrix([[0.983468765322102,-0.0591498262566556,0.171144633832402],[0.0138198967766875,0.966905482105505,0.254760277765986],[-0.180549710852444,-0.248183574654182,0.951738785162699]])
     plots = []
     for i in range(n):
         XP = P*R^i*X
         nextplot = line(XP.transpose(),xmin=-2,xmax=2,ymin=-2,ymax=2,axes=False,aspect_ratio=1)
         plots.append(nextplot)
     a = animate(plots)
     show(a)

     Instead of calculating each element of the rotation matrix (R), I would recommend putting in the formulas in terms of \(u_x\), \(u_y\), \(u_z\), and \(\theta\). You would need to define those variables before they were referenced:

     ux = -0.8137802
     uy = 0.5690534
     uz = 0.1180675
     th = 2*pi/20